A map is continuous if and only if for every set, the image of closure is contained in the closure of image
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As a part of self study, I am trying to prove the following statement:
Suppose $X$ and $Y$ are topological spaces and $f: X rightarrow Y$ is a map. Then $f$ is continuous if and only if $f(overline{A})subseteq overline{f(A)}$, where $overline{A}$ denotes the closure of an arbitrary set $A$.
Assuming $f$ is continuous, the result is almost immediate. Perhaps I am missing something obvious, but I have not been able to make progress on the other direction. Could anyone give me a hint which might illuminate the problem for me?
general-topology continuity
asked Feb 28 '12 at 16:18
Holdsworth88Holdsworth88
5,09232860
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add a comment |
$begingroup$
As a part of self study, I am trying to prove the following statement:
Suppose $X$ and $Y$ are topological spaces and $f: X rightarrow Y$ is a map. Then $f$ is continuous if and only if $f(overline{A})subseteq overline{f(A)}$, where $overline{A}$ denotes the closure of an arbitrary set $A$.
Assuming $f$ is continuous, the result is almost immediate. Perhaps I am missing something obvious, but I have not been able to make progress on the other direction. Could anyone give me a hint which might illuminate the problem for me?
general-topology continuity
asked Feb 28 '12 at 16:18
Holdsworth88Holdsworth88
5,09232860
$endgroup$
$begingroup$
I am not able to come up with any example of a continuous function s.t. $f(overline{A}) subsetneq overline{f(A)}$. Can anyone give such an examples?
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– MUH
Mar 11 '18 at 11:03
2
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Assuming $f$ is continuous, how exactly is the result "immediate"?
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– Al Jebr
May 19 '18 at 18:56
add a comment |
$begingroup$
As a part of self study, I am trying to prove the following statement:
Suppose $X$ and $Y$ are topological spaces and $f: X rightarrow Y$ is a map. Then $f$ is continuous if and only if $f(overline{A})subseteq overline{f(A)}$, where $overline{A}$ denotes the closure of an arbitrary set $A$.
Assuming $f$ is continuous, the result is almost immediate. Perhaps I am missing something obvious, but I have not been able to make progress on the other direction. Could anyone give me a hint which might illuminate the problem for me?
general-topology continuity
asked Feb 28 '12 at 16:18
Holdsworth88Holdsworth88
5,09232860
$endgroup$
As a part of self study, I am trying to prove the following statement:
Suppose $X$ and $Y$ are topological spaces and $f: X rightarrow Y$ is a map. Then $f$ is continuous if and only if $f(overline{A})subseteq overline{f(A)}$, where $overline{A}$ denotes the closure of an arbitrary set $A$.
Assuming $f$ is continuous, the result is almost immediate. Perhaps I am missing something obvious, but I have not been able to make progress on the other direction. Could anyone give me a hint which might illuminate the problem for me?
general-topology continuity
general-topology continuity
asked Feb 28 '12 at 16:18
Holdsworth88Holdsworth88
5,09232860
asked Feb 28 '12 at 16:18
Holdsworth88Holdsworth88
5,09232860
edited Nov 18 '15 at 15:37
user147263
asked Feb 28 '12 at 16:18
Holdsworth88Holdsworth88
5,09232860
asked Feb 28 '12 at 16:18
Holdsworth88Holdsworth88
5,09232860
asked Feb 28 '12 at 16:18
Holdsworth88Holdsworth88
5,09232860
5,09232860
$begingroup$
I am not able to come up with any example of a continuous function s.t. $f(overline{A}) subsetneq overline{f(A)}$. Can anyone give such an examples?
$endgroup$
– MUH
Mar 11 '18 at 11:03
2
$begingroup$
Assuming $f$ is continuous, how exactly is the result "immediate"?
$endgroup$
– Al Jebr
May 19 '18 at 18:56
add a comment |
$begingroup$
I am not able to come up with any example of a continuous function s.t. $f(overline{A}) subsetneq overline{f(A)}$. Can anyone give such an examples?
$endgroup$
– MUH
Mar 11 '18 at 11:03
2
$begingroup$
Assuming $f$ is continuous, how exactly is the result "immediate"?
$endgroup$
– Al Jebr
May 19 '18 at 18:56
$begingroup$
I am not able to come up with any example of a continuous function s.t. $f(overline{A}) subsetneq overline{f(A)}$. Can anyone give such an examples?
$endgroup$
– MUH
Mar 11 '18 at 11:03
$begingroup$
I am not able to come up with any example of a continuous function s.t. $f(overline{A}) subsetneq overline{f(A)}$. Can anyone give such an examples?
$endgroup$
– MUH
Mar 11 '18 at 11:03
2
2
$begingroup$
Assuming $f$ is continuous, how exactly is the result "immediate"?
$endgroup$
– Al Jebr
May 19 '18 at 18:56
$begingroup$
Assuming $f$ is continuous, how exactly is the result "immediate"?
$endgroup$
– Al Jebr
May 19 '18 at 18:56
add a comment |
7 Answers
7
active
oldest
votes
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Because the property is stated in terms of closures, it's I think slightly easier to use the inverse image of closed is closed equivalence of continuity instead:
If $C$ is closed in $Y$, we need to show that $D = f^{-1}[C]$ is closed in $X$.
Now using our closure property for $D$: $$f[overline{D}] subseteq overline{f[D]} = overline{f[f^{-1}[C]]} subseteq overline{C} = C,$$ as $C$ is closed.
This means that $overline{D} subseteq f^{-1}[C] = D$, making $D$ closed, as required.
answered Feb 28 '12 at 20:55
Henno BrandsmaHenno Brandsma
116k349127
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I agree, no need to mess with complements.
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– wildildildlife
Feb 28 '12 at 22:06
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$C$ should be closed in $Y$, not $X$.
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– Holdsworth88
Feb 29 '12 at 10:16
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There's a misleading in the argumentation:<br> What you're showing is that in fact the preimage of closed sets is closed for continuous functions. That is a simple consequence as: $f^{-1}(C)=f^{-1}(C)^{c c}=f^{-1}(C^c)^c text{closed}$. However, the assertion is much more subtle!<br> Moreover there's a missing step in your derivation, which is precisely the desired assertion: $f(overline{D})subseteq overline{f(D)}text{???}$
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– C-Star-W-Star
Jan 22 '14 at 16:57
1
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@Freeze_S, it seems to me that you have misread the question. The OP was looking for the direction that Henno has proved here.
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– Carsten S
Jan 22 '14 at 18:09
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why the result is imeadiat if $f$ is continuous?
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– Guerlando OCs
May 9 '16 at 0:18
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show 5 more comments
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If $f$ is continuous, then $f^{-1}(Y-overline{f(A)})$ is open; since $f^{-1}(Y-overline{f(A)}) = X -f^{-1}(overline{f(A)})$, then $f^{-1}(overline{f(A)})$ is closed; since $Asubseteq f^{-1}(f(A))subseteq f^{-1}overline{f(A)}$, we conclude that $overline{A}subseteq f^{-1}(overline{f(A)})$, and therefore $fleft(overline{A}right)subseteq fleft(f^{-1}left(overline{f(A)}right)right)subseteq overline{f(A)}$. This proves one direction. (Since you said you already had this, I posted the full argument).
Conversely, assume that for every $A$, $fleft(overline{A}right)subseteq overline{f(A)}$. Let $V$ be an open subset of $Y$; we want to show that $f^{-1}(V)$ is open; equivalently, we want to show that $X-f^{-1}(V)$ is closed; that is, we want to show that $X-f^{-1}(V)=overline{X-f^{-1}(V)}$.
By assumption, $fleft(overline{X-f^{-1}(V)}right)subseteq overline{f(X-f^{-1}(V))}$. Now, $X-f^{-1}(V) = f^{-1}(Y-V)$, so $f(X-f^{-1}(V)) subseteq Y-V$, which is closed; so
$$fleft(overline{X-f^{-1}(V)}right)=fleft(overline{f^{-1}(Y-V)}right)subseteqoverline{fleft(f^{-1}(Y-V)right)}subseteq Y-V.$$ Therefore,
$$overline{X-f^{-1}(V)} subseteq f^{-1}(Y-V).$$
Is this sufficient?
answered Feb 28 '12 at 16:26
Arturo MagidinArturo Magidin
266k34591922
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That was more than sufficient, thank you.
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– Holdsworth88
Feb 28 '12 at 16:51
1
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Converse part is nice.
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– math is love
Jan 26 '17 at 15:15
add a comment |
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We show that $f$ is continuous at each $xin X$. Recall that $f$ is continuous at $x$ if for any open nhood $V$ of $f(x)$, there is an open nhood $U$ of $x$ with $f(U)subseteq V$.
So, let $xin X$ and let $V$ be an open nhood of $f(x)$.
Set $E=Xsetminus f^{-1}(V)$. Noting that $f(E)$ is contained in the closed set $V^C$,
we see that
$$f(overline{E})subseteq overline{f(E)}subseteq V^C; $$
whence $xnotin overline E$ (since $f(x)$ is in $V$).
But then $x$ is in the open set $Xsetminus overline{E}$. Moreover, since $Xsetminus overline{E}subseteq Xsetminus E=f^{-1}(V)$, it follows that $f(Xsetminus overline{E})subseteq V$, as desired.
An aside:
If $X$ were first countable, we could argue using sequences (where a sequence $(x_n)$ converges to $x$ if for any nhood $U$ of $x$ there is an $N$ so that $x_min U$ for all $mge N$); for in such a space a function is continuous at $x$ if and only if $(f (x_n))$ converges to $f(x)$ whenever $x_n$ converges to $x$.
Indeed, it is easily verified that given $x_nrightarrow x$ and any subsequence $(x_{n_k})$ of $(x_n)$, that the image of this subsequence under $f$ when thought of as a sequence has a subsequence that converges to $f(x)$. Thus every subsequence of $(f(x_n))$ has a further subsequence which converges to $f(x)$, which implies that $(f(x_n))$ converges to $x$.
Of course, $X$ need not be first countable...
Though I think the sequential method above is a bit much here, I wonder if the argument can be suitably modified to arbitrary $X$ by using nets?
answered Feb 28 '12 at 18:10
David MitraDavid Mitra
63.8k6102165
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Henno’s argument is probably the slickest, but one can also work pointwise, either directly or, if one already has those characterizations of continuity, with nets or filters.
Suppose that $f$ is not continuous. Then there is some point $x_0in X$ at which $f$ fails to be continuous. If $y_0=f(x_0)$, this means that there is an open nbhd $U$ of $y_0$ such that if $V$ is an open nbhd of $x_0$, there is a point $x_Vin V$ such that $f(x_V)notin U$. Let $$A={x_V:Vtext{ is an open nbhd of }x_0};.$$
Clearly $x_0inoverline{A}$, and $f[A]subseteq Ysetminus U$. Moreover, $Ysetminus U$ is closed, so $overline{f[A]}subseteq Ysetminus U$, and hence
$$y_0in f[overline{A}]subseteqoverline{f[A]}subseteq Ysetminus U;,$$
contradicting the choice of $U$.
Those who like nets may notice that $A$ actually is a net, over the directed set $langlemathscr{N}(x_0),supseteqrangle$, where $mathscr{N}(x_0)$ is the family of open nbhds of $x_0$, and the argument shows that $f$ doesn’t preserve the limit of this net. (I could of course just as well have used the nbhd filter at $x_0$, instead of using only open nbhds.)
Those who prefer filters may modify this to consider the filter $$mathscr{F}={Vcap f^{-1}[Ysetminus U]:Vinmathscr{N}(x_0)}$$ instead.
answered Feb 29 '12 at 6:26
Brian M. ScottBrian M. Scott
461k40518920
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Here's one proof of the converse provided $X$ and $Y$ are metric spaces:
Take a limit point $x$ of $A$. Then because $f(overline{A}) subseteq overline{f(A)}$, we have that $f(x)$ is a limit point of $f(A)$.
Because $x$ is a limit point of $A$, for every $delta > 0$ there is a point $p in A$ with $|x - p| < delta$. And because $f(x)$ is a limit point of $f(A)$, for every $epsilon > 0$, there is a point $q in f(A)$ with $|f(x) - f(p)| < epsilon$.
We thus have that for every point $x in A$ with $|x - p| < delta$, in fact $|f(x) - f(p)| < epsilon$, so $f$ must be continuous.
edited Feb 28 '12 at 17:11
Austin Mohr
20.9k35299
answered Feb 28 '12 at 16:33
jamaicanwormjamaicanworm
1,17762638
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I guess this is a decent enough answer for the case of the real numbers, but the question was about arbitrary topological spaces $X$ and $Y$.
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– kahen
Feb 28 '12 at 16:45
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We are not in a world where things like $|x-p|$ could make sense. Any "TOPOLOGICAL SPACE" is in question, but any way a good proof that will be useful. I am upvoting and don't delete this answer because of this as it will help close duplicates.
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– user21436
Feb 28 '12 at 16:46
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Can't you just substitute $d(x, p)$ for $|x - p|$?
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– jamaicanworm
Feb 28 '12 at 16:50
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@jamaicaworm: Only in a metric space. Not every topological space is a metric space, or even metrizable.
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– Arturo Magidin
Feb 28 '12 at 16:50
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@jamaicanworm: The post is labeled [general-topology], not [metric-spaces]. The definition of a topological space includes the notion of "open set", and "closed set" is a set whose complement is open. See the definition in Wikipedia.
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– Arturo Magidin
Feb 28 '12 at 16:59
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A proof using nets:
Suppose $yin f(overline{A})backslash overline{fA}$. Taking $xin overline{A}$ with $y=f(x)$; there is a net $x_iin A$ converging to $x$. Now $f(x_i)$ is a net in $f(A)$ which does not converge to $f(x)$ [for this would imply $f(x)in overline{fA}$]. Thus $f$ is not continuous.
answered Feb 28 '12 at 22:02
wildildildlifewildildildlife
4,5142222
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This is the direction that the OP had already been able to show.
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– Brian M. Scott
Feb 29 '12 at 6:08
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@Brian: Yes, sorry! I realized this the day after...
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– wildildildlife
Mar 2 '12 at 13:15
add a comment |
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The assertion is equivalent to:
$overline{A}subseteq f^{-1}(overline{f(A)})$
So, the assertion follows from:
$overline{A}subseteqoverline{f^{-1}(f(A))}subseteqoverline{f^{-1}(overline{f(A)})}=f^{-1}(overline{f(A)})$
- Inclusion: $Asubseteq f^{-1}(f(A)) Rightarrow overline{A}subseteqoverline{f^{-1}(f(A))}$
- Inclusion: $f(A)subseteqoverline{f(A)} Rightarrow f^{-1}(f(A))subseteq f^{-1}(overline{f(A)}) Rightarrow overline{f^{-1}(f(A))}subseteq overline{f^{-1}(overline{f(A)})}$
- Equality: $overline{f(A)} text{ closed} Rightarrow f^{-1}(overline{f(A)}) text{ closed} Rightarrow overline{f^{-1}(overline{f(A)})}=f^{-1}(overline{f(A)})$
The converse assertion is equivalent to:
$overline{B}=B Rightarrow overline{f^{-1}(B)}=f^{-1}(B)$
So, the converse assertion follows from:
$f^{-1}(B)subseteqoverline{f^{-1}(B)}subseteq f^{-1}(f(overline{f^{-1}(B)}))subseteq f^{-1}(overline{f(f^{-1}(B))}) subseteq f^{-1}(overline{B}) =f^{-1}(B)$
That gives:
$f^{-1}(B)=overline{f^{-1}(B)}$
- Inclusion: $Asubseteq overline{A} text{ in general}$
- Inclusion: $Asubseteq f^{-1}(f(A)) text{ in general}$
- Inclusion: $f(overline{A})subseteq overline{f(A)} text{ by assumption}$
- Inclusion: $f(f^{-1}(B))subseteq B text{ in general} Rightarrow overline{f(f^{-1}(B))}subseteq overline{B} Rightarrow f^{-1}(overline{f(f^{-1}(B))})subseteq f^{-1}(overline{B})$
- Equality: $overline{B}=B Rightarrow f^{-1}(overline{B})=f^{-1}(B)$
answered Jan 22 '14 at 17:38
C-Star-W-StarC-Star-W-Star
8,36642166
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I don't see how this is shorter than Henno's answer, which is complete since the OP cleared out the other direction. I don't think it is good practice to claim your answer is somehow better than others, and specifically target another user's answer.
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– Pedro Tamaroff♦
Jan 22 '14 at 17:51
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I'm sorry for that ...got a little bit upset... I deleted that part so there's no bad discussion about it -sorry for that! Anyway Henno only shows that the preimage of a closed sst is closed what is a rather simple consequence. But the assertion is much more tricky than that. See my comment on Henno's answer.
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– C-Star-W-Star
Jan 22 '14 at 17:57
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7 Answers
7
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7 Answers
7
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$begingroup$
Because the property is stated in terms of closures, it's I think slightly easier to use the inverse image of closed is closed equivalence of continuity instead:
If $C$ is closed in $Y$, we need to show that $D = f^{-1}[C]$ is closed in $X$.
Now using our closure property for $D$: $$f[overline{D}] subseteq overline{f[D]} = overline{f[f^{-1}[C]]} subseteq overline{C} = C,$$ as $C$ is closed.
This means that $overline{D} subseteq f^{-1}[C] = D$, making $D$ closed, as required.
answered Feb 28 '12 at 20:55
Henno BrandsmaHenno Brandsma
116k349127
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I agree, no need to mess with complements.
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– wildildildlife
Feb 28 '12 at 22:06
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$C$ should be closed in $Y$, not $X$.
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– Holdsworth88
Feb 29 '12 at 10:16
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There's a misleading in the argumentation:<br> What you're showing is that in fact the preimage of closed sets is closed for continuous functions. That is a simple consequence as: $f^{-1}(C)=f^{-1}(C)^{c c}=f^{-1}(C^c)^c text{closed}$. However, the assertion is much more subtle!<br> Moreover there's a missing step in your derivation, which is precisely the desired assertion: $f(overline{D})subseteq overline{f(D)}text{???}$
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– C-Star-W-Star
Jan 22 '14 at 16:57
1
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@Freeze_S, it seems to me that you have misread the question. The OP was looking for the direction that Henno has proved here.
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– Carsten S
Jan 22 '14 at 18:09
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why the result is imeadiat if $f$ is continuous?
$endgroup$
– Guerlando OCs
May 9 '16 at 0:18
|
show 5 more comments
$begingroup$
Because the property is stated in terms of closures, it's I think slightly easier to use the inverse image of closed is closed equivalence of continuity instead:
If $C$ is closed in $Y$, we need to show that $D = f^{-1}[C]$ is closed in $X$.
Now using our closure property for $D$: $$f[overline{D}] subseteq overline{f[D]} = overline{f[f^{-1}[C]]} subseteq overline{C} = C,$$ as $C$ is closed.
This means that $overline{D} subseteq f^{-1}[C] = D$, making $D$ closed, as required.
answered Feb 28 '12 at 20:55
Henno BrandsmaHenno Brandsma
116k349127
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$begingroup$
I agree, no need to mess with complements.
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– wildildildlife
Feb 28 '12 at 22:06
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$C$ should be closed in $Y$, not $X$.
$endgroup$
– Holdsworth88
Feb 29 '12 at 10:16
$begingroup$
There's a misleading in the argumentation:<br> What you're showing is that in fact the preimage of closed sets is closed for continuous functions. That is a simple consequence as: $f^{-1}(C)=f^{-1}(C)^{c c}=f^{-1}(C^c)^c text{closed}$. However, the assertion is much more subtle!<br> Moreover there's a missing step in your derivation, which is precisely the desired assertion: $f(overline{D})subseteq overline{f(D)}text{???}$
$endgroup$
– C-Star-W-Star
Jan 22 '14 at 16:57
1
$begingroup$
@Freeze_S, it seems to me that you have misread the question. The OP was looking for the direction that Henno has proved here.
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– Carsten S
Jan 22 '14 at 18:09
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why the result is imeadiat if $f$ is continuous?
$endgroup$
– Guerlando OCs
May 9 '16 at 0:18
|
show 5 more comments
$begingroup$
Because the property is stated in terms of closures, it's I think slightly easier to use the inverse image of closed is closed equivalence of continuity instead:
If $C$ is closed in $Y$, we need to show that $D = f^{-1}[C]$ is closed in $X$.
Now using our closure property for $D$: $$f[overline{D}] subseteq overline{f[D]} = overline{f[f^{-1}[C]]} subseteq overline{C} = C,$$ as $C$ is closed.
This means that $overline{D} subseteq f^{-1}[C] = D$, making $D$ closed, as required.
answered Feb 28 '12 at 20:55
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
Because the property is stated in terms of closures, it's I think slightly easier to use the inverse image of closed is closed equivalence of continuity instead:
If $C$ is closed in $Y$, we need to show that $D = f^{-1}[C]$ is closed in $X$.
Now using our closure property for $D$: $$f[overline{D}] subseteq overline{f[D]} = overline{f[f^{-1}[C]]} subseteq overline{C} = C,$$ as $C$ is closed.
This means that $overline{D} subseteq f^{-1}[C] = D$, making $D$ closed, as required.
answered Feb 28 '12 at 20:55
Henno BrandsmaHenno Brandsma
116k349127
edited Nov 17 '18 at 9:15
answered Feb 28 '12 at 20:55
Henno BrandsmaHenno Brandsma
116k349127
answered Feb 28 '12 at 20:55
Henno BrandsmaHenno Brandsma
116k349127
answered Feb 28 '12 at 20:55
Henno BrandsmaHenno Brandsma
116k349127
116k349127
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I agree, no need to mess with complements.
$endgroup$
– wildildildlife
Feb 28 '12 at 22:06
$begingroup$
$C$ should be closed in $Y$, not $X$.
$endgroup$
– Holdsworth88
Feb 29 '12 at 10:16
$begingroup$
There's a misleading in the argumentation:<br> What you're showing is that in fact the preimage of closed sets is closed for continuous functions. That is a simple consequence as: $f^{-1}(C)=f^{-1}(C)^{c c}=f^{-1}(C^c)^c text{closed}$. However, the assertion is much more subtle!<br> Moreover there's a missing step in your derivation, which is precisely the desired assertion: $f(overline{D})subseteq overline{f(D)}text{???}$
$endgroup$
– C-Star-W-Star
Jan 22 '14 at 16:57
1
$begingroup$
@Freeze_S, it seems to me that you have misread the question. The OP was looking for the direction that Henno has proved here.
$endgroup$
– Carsten S
Jan 22 '14 at 18:09
$begingroup$
why the result is imeadiat if $f$ is continuous?
$endgroup$
– Guerlando OCs
May 9 '16 at 0:18
|
show 5 more comments
$begingroup$
I agree, no need to mess with complements.
$endgroup$
– wildildildlife
Feb 28 '12 at 22:06
$begingroup$
$C$ should be closed in $Y$, not $X$.
$endgroup$
– Holdsworth88
Feb 29 '12 at 10:16
$begingroup$
There's a misleading in the argumentation:<br> What you're showing is that in fact the preimage of closed sets is closed for continuous functions. That is a simple consequence as: $f^{-1}(C)=f^{-1}(C)^{c c}=f^{-1}(C^c)^c text{closed}$. However, the assertion is much more subtle!<br> Moreover there's a missing step in your derivation, which is precisely the desired assertion: $f(overline{D})subseteq overline{f(D)}text{???}$
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– C-Star-W-Star
Jan 22 '14 at 16:57
1
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@Freeze_S, it seems to me that you have misread the question. The OP was looking for the direction that Henno has proved here.
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– Carsten S
Jan 22 '14 at 18:09
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why the result is imeadiat if $f$ is continuous?
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– Guerlando OCs
May 9 '16 at 0:18
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I agree, no need to mess with complements.
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– wildildildlife
Feb 28 '12 at 22:06
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I agree, no need to mess with complements.
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– wildildildlife
Feb 28 '12 at 22:06
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$C$ should be closed in $Y$, not $X$.
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– Holdsworth88
Feb 29 '12 at 10:16
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$C$ should be closed in $Y$, not $X$.
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– Holdsworth88
Feb 29 '12 at 10:16
$begingroup$
There's a misleading in the argumentation:<br> What you're showing is that in fact the preimage of closed sets is closed for continuous functions. That is a simple consequence as: $f^{-1}(C)=f^{-1}(C)^{c c}=f^{-1}(C^c)^c text{closed}$. However, the assertion is much more subtle!<br> Moreover there's a missing step in your derivation, which is precisely the desired assertion: $f(overline{D})subseteq overline{f(D)}text{???}$
$endgroup$
– C-Star-W-Star
Jan 22 '14 at 16:57
$begingroup$
There's a misleading in the argumentation:<br> What you're showing is that in fact the preimage of closed sets is closed for continuous functions. That is a simple consequence as: $f^{-1}(C)=f^{-1}(C)^{c c}=f^{-1}(C^c)^c text{closed}$. However, the assertion is much more subtle!<br> Moreover there's a missing step in your derivation, which is precisely the desired assertion: $f(overline{D})subseteq overline{f(D)}text{???}$
$endgroup$
– C-Star-W-Star
Jan 22 '14 at 16:57
1
1
$begingroup$
@Freeze_S, it seems to me that you have misread the question. The OP was looking for the direction that Henno has proved here.
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– Carsten S
Jan 22 '14 at 18:09
$begingroup$
@Freeze_S, it seems to me that you have misread the question. The OP was looking for the direction that Henno has proved here.
$endgroup$
– Carsten S
Jan 22 '14 at 18:09
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why the result is imeadiat if $f$ is continuous?
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– Guerlando OCs
May 9 '16 at 0:18
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why the result is imeadiat if $f$ is continuous?
$endgroup$
– Guerlando OCs
May 9 '16 at 0:18
|
show 5 more comments
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If $f$ is continuous, then $f^{-1}(Y-overline{f(A)})$ is open; since $f^{-1}(Y-overline{f(A)}) = X -f^{-1}(overline{f(A)})$, then $f^{-1}(overline{f(A)})$ is closed; since $Asubseteq f^{-1}(f(A))subseteq f^{-1}overline{f(A)}$, we conclude that $overline{A}subseteq f^{-1}(overline{f(A)})$, and therefore $fleft(overline{A}right)subseteq fleft(f^{-1}left(overline{f(A)}right)right)subseteq overline{f(A)}$. This proves one direction. (Since you said you already had this, I posted the full argument).
Conversely, assume that for every $A$, $fleft(overline{A}right)subseteq overline{f(A)}$. Let $V$ be an open subset of $Y$; we want to show that $f^{-1}(V)$ is open; equivalently, we want to show that $X-f^{-1}(V)$ is closed; that is, we want to show that $X-f^{-1}(V)=overline{X-f^{-1}(V)}$.
By assumption, $fleft(overline{X-f^{-1}(V)}right)subseteq overline{f(X-f^{-1}(V))}$. Now, $X-f^{-1}(V) = f^{-1}(Y-V)$, so $f(X-f^{-1}(V)) subseteq Y-V$, which is closed; so
$$fleft(overline{X-f^{-1}(V)}right)=fleft(overline{f^{-1}(Y-V)}right)subseteqoverline{fleft(f^{-1}(Y-V)right)}subseteq Y-V.$$ Therefore,
$$overline{X-f^{-1}(V)} subseteq f^{-1}(Y-V).$$
Is this sufficient?
answered Feb 28 '12 at 16:26
Arturo MagidinArturo Magidin
266k34591922
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2
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That was more than sufficient, thank you.
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– Holdsworth88
Feb 28 '12 at 16:51
1
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Converse part is nice.
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– math is love
Jan 26 '17 at 15:15
add a comment |
$begingroup$
If $f$ is continuous, then $f^{-1}(Y-overline{f(A)})$ is open; since $f^{-1}(Y-overline{f(A)}) = X -f^{-1}(overline{f(A)})$, then $f^{-1}(overline{f(A)})$ is closed; since $Asubseteq f^{-1}(f(A))subseteq f^{-1}overline{f(A)}$, we conclude that $overline{A}subseteq f^{-1}(overline{f(A)})$, and therefore $fleft(overline{A}right)subseteq fleft(f^{-1}left(overline{f(A)}right)right)subseteq overline{f(A)}$. This proves one direction. (Since you said you already had this, I posted the full argument).
Conversely, assume that for every $A$, $fleft(overline{A}right)subseteq overline{f(A)}$. Let $V$ be an open subset of $Y$; we want to show that $f^{-1}(V)$ is open; equivalently, we want to show that $X-f^{-1}(V)$ is closed; that is, we want to show that $X-f^{-1}(V)=overline{X-f^{-1}(V)}$.
By assumption, $fleft(overline{X-f^{-1}(V)}right)subseteq overline{f(X-f^{-1}(V))}$. Now, $X-f^{-1}(V) = f^{-1}(Y-V)$, so $f(X-f^{-1}(V)) subseteq Y-V$, which is closed; so
$$fleft(overline{X-f^{-1}(V)}right)=fleft(overline{f^{-1}(Y-V)}right)subseteqoverline{fleft(f^{-1}(Y-V)right)}subseteq Y-V.$$ Therefore,
$$overline{X-f^{-1}(V)} subseteq f^{-1}(Y-V).$$
Is this sufficient?
answered Feb 28 '12 at 16:26
Arturo MagidinArturo Magidin
266k34591922
$endgroup$
2
$begingroup$
That was more than sufficient, thank you.
$endgroup$
– Holdsworth88
Feb 28 '12 at 16:51
1
$begingroup$
Converse part is nice.
$endgroup$
– math is love
Jan 26 '17 at 15:15
add a comment |
$begingroup$
If $f$ is continuous, then $f^{-1}(Y-overline{f(A)})$ is open; since $f^{-1}(Y-overline{f(A)}) = X -f^{-1}(overline{f(A)})$, then $f^{-1}(overline{f(A)})$ is closed; since $Asubseteq f^{-1}(f(A))subseteq f^{-1}overline{f(A)}$, we conclude that $overline{A}subseteq f^{-1}(overline{f(A)})$, and therefore $fleft(overline{A}right)subseteq fleft(f^{-1}left(overline{f(A)}right)right)subseteq overline{f(A)}$. This proves one direction. (Since you said you already had this, I posted the full argument).
Conversely, assume that for every $A$, $fleft(overline{A}right)subseteq overline{f(A)}$. Let $V$ be an open subset of $Y$; we want to show that $f^{-1}(V)$ is open; equivalently, we want to show that $X-f^{-1}(V)$ is closed; that is, we want to show that $X-f^{-1}(V)=overline{X-f^{-1}(V)}$.
By assumption, $fleft(overline{X-f^{-1}(V)}right)subseteq overline{f(X-f^{-1}(V))}$. Now, $X-f^{-1}(V) = f^{-1}(Y-V)$, so $f(X-f^{-1}(V)) subseteq Y-V$, which is closed; so
$$fleft(overline{X-f^{-1}(V)}right)=fleft(overline{f^{-1}(Y-V)}right)subseteqoverline{fleft(f^{-1}(Y-V)right)}subseteq Y-V.$$ Therefore,
$$overline{X-f^{-1}(V)} subseteq f^{-1}(Y-V).$$
Is this sufficient?
answered Feb 28 '12 at 16:26
Arturo MagidinArturo Magidin
266k34591922
$endgroup$
If $f$ is continuous, then $f^{-1}(Y-overline{f(A)})$ is open; since $f^{-1}(Y-overline{f(A)}) = X -f^{-1}(overline{f(A)})$, then $f^{-1}(overline{f(A)})$ is closed; since $Asubseteq f^{-1}(f(A))subseteq f^{-1}overline{f(A)}$, we conclude that $overline{A}subseteq f^{-1}(overline{f(A)})$, and therefore $fleft(overline{A}right)subseteq fleft(f^{-1}left(overline{f(A)}right)right)subseteq overline{f(A)}$. This proves one direction. (Since you said you already had this, I posted the full argument).
Conversely, assume that for every $A$, $fleft(overline{A}right)subseteq overline{f(A)}$. Let $V$ be an open subset of $Y$; we want to show that $f^{-1}(V)$ is open; equivalently, we want to show that $X-f^{-1}(V)$ is closed; that is, we want to show that $X-f^{-1}(V)=overline{X-f^{-1}(V)}$.
By assumption, $fleft(overline{X-f^{-1}(V)}right)subseteq overline{f(X-f^{-1}(V))}$. Now, $X-f^{-1}(V) = f^{-1}(Y-V)$, so $f(X-f^{-1}(V)) subseteq Y-V$, which is closed; so
$$fleft(overline{X-f^{-1}(V)}right)=fleft(overline{f^{-1}(Y-V)}right)subseteqoverline{fleft(f^{-1}(Y-V)right)}subseteq Y-V.$$ Therefore,
$$overline{X-f^{-1}(V)} subseteq f^{-1}(Y-V).$$
Is this sufficient?
answered Feb 28 '12 at 16:26
Arturo MagidinArturo Magidin
266k34591922
edited Feb 28 '12 at 17:01
answered Feb 28 '12 at 16:26
Arturo MagidinArturo Magidin
266k34591922
answered Feb 28 '12 at 16:26
Arturo MagidinArturo Magidin
266k34591922
answered Feb 28 '12 at 16:26
Arturo MagidinArturo Magidin
266k34591922
266k34591922
2
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That was more than sufficient, thank you.
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– Holdsworth88
Feb 28 '12 at 16:51
1
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Converse part is nice.
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– math is love
Jan 26 '17 at 15:15
add a comment |
2
$begingroup$
That was more than sufficient, thank you.
$endgroup$
– Holdsworth88
Feb 28 '12 at 16:51
1
$begingroup$
Converse part is nice.
$endgroup$
– math is love
Jan 26 '17 at 15:15
2
2
$begingroup$
That was more than sufficient, thank you.
$endgroup$
– Holdsworth88
Feb 28 '12 at 16:51
$begingroup$
That was more than sufficient, thank you.
$endgroup$
– Holdsworth88
Feb 28 '12 at 16:51
1
1
$begingroup$
Converse part is nice.
$endgroup$
– math is love
Jan 26 '17 at 15:15
$begingroup$
Converse part is nice.
$endgroup$
– math is love
Jan 26 '17 at 15:15
add a comment |
$begingroup$
We show that $f$ is continuous at each $xin X$. Recall that $f$ is continuous at $x$ if for any open nhood $V$ of $f(x)$, there is an open nhood $U$ of $x$ with $f(U)subseteq V$.
So, let $xin X$ and let $V$ be an open nhood of $f(x)$.
Set $E=Xsetminus f^{-1}(V)$. Noting that $f(E)$ is contained in the closed set $V^C$,
we see that
$$f(overline{E})subseteq overline{f(E)}subseteq V^C; $$
whence $xnotin overline E$ (since $f(x)$ is in $V$).
But then $x$ is in the open set $Xsetminus overline{E}$. Moreover, since $Xsetminus overline{E}subseteq Xsetminus E=f^{-1}(V)$, it follows that $f(Xsetminus overline{E})subseteq V$, as desired.
An aside:
If $X$ were first countable, we could argue using sequences (where a sequence $(x_n)$ converges to $x$ if for any nhood $U$ of $x$ there is an $N$ so that $x_min U$ for all $mge N$); for in such a space a function is continuous at $x$ if and only if $(f (x_n))$ converges to $f(x)$ whenever $x_n$ converges to $x$.
Indeed, it is easily verified that given $x_nrightarrow x$ and any subsequence $(x_{n_k})$ of $(x_n)$, that the image of this subsequence under $f$ when thought of as a sequence has a subsequence that converges to $f(x)$. Thus every subsequence of $(f(x_n))$ has a further subsequence which converges to $f(x)$, which implies that $(f(x_n))$ converges to $x$.
Of course, $X$ need not be first countable...
Though I think the sequential method above is a bit much here, I wonder if the argument can be suitably modified to arbitrary $X$ by using nets?
answered Feb 28 '12 at 18:10
David MitraDavid Mitra
63.8k6102165
$endgroup$
add a comment |
$begingroup$
We show that $f$ is continuous at each $xin X$. Recall that $f$ is continuous at $x$ if for any open nhood $V$ of $f(x)$, there is an open nhood $U$ of $x$ with $f(U)subseteq V$.
So, let $xin X$ and let $V$ be an open nhood of $f(x)$.
Set $E=Xsetminus f^{-1}(V)$. Noting that $f(E)$ is contained in the closed set $V^C$,
we see that
$$f(overline{E})subseteq overline{f(E)}subseteq V^C; $$
whence $xnotin overline E$ (since $f(x)$ is in $V$).
But then $x$ is in the open set $Xsetminus overline{E}$. Moreover, since $Xsetminus overline{E}subseteq Xsetminus E=f^{-1}(V)$, it follows that $f(Xsetminus overline{E})subseteq V$, as desired.
An aside:
If $X$ were first countable, we could argue using sequences (where a sequence $(x_n)$ converges to $x$ if for any nhood $U$ of $x$ there is an $N$ so that $x_min U$ for all $mge N$); for in such a space a function is continuous at $x$ if and only if $(f (x_n))$ converges to $f(x)$ whenever $x_n$ converges to $x$.
Indeed, it is easily verified that given $x_nrightarrow x$ and any subsequence $(x_{n_k})$ of $(x_n)$, that the image of this subsequence under $f$ when thought of as a sequence has a subsequence that converges to $f(x)$. Thus every subsequence of $(f(x_n))$ has a further subsequence which converges to $f(x)$, which implies that $(f(x_n))$ converges to $x$.
Of course, $X$ need not be first countable...
Though I think the sequential method above is a bit much here, I wonder if the argument can be suitably modified to arbitrary $X$ by using nets?
answered Feb 28 '12 at 18:10
David MitraDavid Mitra
63.8k6102165
$endgroup$
add a comment |
$begingroup$
We show that $f$ is continuous at each $xin X$. Recall that $f$ is continuous at $x$ if for any open nhood $V$ of $f(x)$, there is an open nhood $U$ of $x$ with $f(U)subseteq V$.
So, let $xin X$ and let $V$ be an open nhood of $f(x)$.
Set $E=Xsetminus f^{-1}(V)$. Noting that $f(E)$ is contained in the closed set $V^C$,
we see that
$$f(overline{E})subseteq overline{f(E)}subseteq V^C; $$
whence $xnotin overline E$ (since $f(x)$ is in $V$).
But then $x$ is in the open set $Xsetminus overline{E}$. Moreover, since $Xsetminus overline{E}subseteq Xsetminus E=f^{-1}(V)$, it follows that $f(Xsetminus overline{E})subseteq V$, as desired.
An aside:
If $X$ were first countable, we could argue using sequences (where a sequence $(x_n)$ converges to $x$ if for any nhood $U$ of $x$ there is an $N$ so that $x_min U$ for all $mge N$); for in such a space a function is continuous at $x$ if and only if $(f (x_n))$ converges to $f(x)$ whenever $x_n$ converges to $x$.
Indeed, it is easily verified that given $x_nrightarrow x$ and any subsequence $(x_{n_k})$ of $(x_n)$, that the image of this subsequence under $f$ when thought of as a sequence has a subsequence that converges to $f(x)$. Thus every subsequence of $(f(x_n))$ has a further subsequence which converges to $f(x)$, which implies that $(f(x_n))$ converges to $x$.
Of course, $X$ need not be first countable...
Though I think the sequential method above is a bit much here, I wonder if the argument can be suitably modified to arbitrary $X$ by using nets?
answered Feb 28 '12 at 18:10
David MitraDavid Mitra
63.8k6102165
$endgroup$
We show that $f$ is continuous at each $xin X$. Recall that $f$ is continuous at $x$ if for any open nhood $V$ of $f(x)$, there is an open nhood $U$ of $x$ with $f(U)subseteq V$.
So, let $xin X$ and let $V$ be an open nhood of $f(x)$.
Set $E=Xsetminus f^{-1}(V)$. Noting that $f(E)$ is contained in the closed set $V^C$,
we see that
$$f(overline{E})subseteq overline{f(E)}subseteq V^C; $$
whence $xnotin overline E$ (since $f(x)$ is in $V$).
But then $x$ is in the open set $Xsetminus overline{E}$. Moreover, since $Xsetminus overline{E}subseteq Xsetminus E=f^{-1}(V)$, it follows that $f(Xsetminus overline{E})subseteq V$, as desired.
An aside:
If $X$ were first countable, we could argue using sequences (where a sequence $(x_n)$ converges to $x$ if for any nhood $U$ of $x$ there is an $N$ so that $x_min U$ for all $mge N$); for in such a space a function is continuous at $x$ if and only if $(f (x_n))$ converges to $f(x)$ whenever $x_n$ converges to $x$.
Indeed, it is easily verified that given $x_nrightarrow x$ and any subsequence $(x_{n_k})$ of $(x_n)$, that the image of this subsequence under $f$ when thought of as a sequence has a subsequence that converges to $f(x)$. Thus every subsequence of $(f(x_n))$ has a further subsequence which converges to $f(x)$, which implies that $(f(x_n))$ converges to $x$.
Of course, $X$ need not be first countable...
Though I think the sequential method above is a bit much here, I wonder if the argument can be suitably modified to arbitrary $X$ by using nets?
answered Feb 28 '12 at 18:10
David MitraDavid Mitra
63.8k6102165
answered Feb 28 '12 at 18:10
David MitraDavid Mitra
63.8k6102165
answered Feb 28 '12 at 18:10
David MitraDavid Mitra
63.8k6102165
answered Feb 28 '12 at 18:10
David MitraDavid Mitra
63.8k6102165
63.8k6102165
add a comment |
add a comment |
$begingroup$
Henno’s argument is probably the slickest, but one can also work pointwise, either directly or, if one already has those characterizations of continuity, with nets or filters.
Suppose that $f$ is not continuous. Then there is some point $x_0in X$ at which $f$ fails to be continuous. If $y_0=f(x_0)$, this means that there is an open nbhd $U$ of $y_0$ such that if $V$ is an open nbhd of $x_0$, there is a point $x_Vin V$ such that $f(x_V)notin U$. Let $$A={x_V:Vtext{ is an open nbhd of }x_0};.$$
Clearly $x_0inoverline{A}$, and $f[A]subseteq Ysetminus U$. Moreover, $Ysetminus U$ is closed, so $overline{f[A]}subseteq Ysetminus U$, and hence
$$y_0in f[overline{A}]subseteqoverline{f[A]}subseteq Ysetminus U;,$$
contradicting the choice of $U$.
Those who like nets may notice that $A$ actually is a net, over the directed set $langlemathscr{N}(x_0),supseteqrangle$, where $mathscr{N}(x_0)$ is the family of open nbhds of $x_0$, and the argument shows that $f$ doesn’t preserve the limit of this net. (I could of course just as well have used the nbhd filter at $x_0$, instead of using only open nbhds.)
Those who prefer filters may modify this to consider the filter $$mathscr{F}={Vcap f^{-1}[Ysetminus U]:Vinmathscr{N}(x_0)}$$ instead.
answered Feb 29 '12 at 6:26
Brian M. ScottBrian M. Scott
461k40518920
$endgroup$
add a comment |
$begingroup$
Henno’s argument is probably the slickest, but one can also work pointwise, either directly or, if one already has those characterizations of continuity, with nets or filters.
Suppose that $f$ is not continuous. Then there is some point $x_0in X$ at which $f$ fails to be continuous. If $y_0=f(x_0)$, this means that there is an open nbhd $U$ of $y_0$ such that if $V$ is an open nbhd of $x_0$, there is a point $x_Vin V$ such that $f(x_V)notin U$. Let $$A={x_V:Vtext{ is an open nbhd of }x_0};.$$
Clearly $x_0inoverline{A}$, and $f[A]subseteq Ysetminus U$. Moreover, $Ysetminus U$ is closed, so $overline{f[A]}subseteq Ysetminus U$, and hence
$$y_0in f[overline{A}]subseteqoverline{f[A]}subseteq Ysetminus U;,$$
contradicting the choice of $U$.
Those who like nets may notice that $A$ actually is a net, over the directed set $langlemathscr{N}(x_0),supseteqrangle$, where $mathscr{N}(x_0)$ is the family of open nbhds of $x_0$, and the argument shows that $f$ doesn’t preserve the limit of this net. (I could of course just as well have used the nbhd filter at $x_0$, instead of using only open nbhds.)
Those who prefer filters may modify this to consider the filter $$mathscr{F}={Vcap f^{-1}[Ysetminus U]:Vinmathscr{N}(x_0)}$$ instead.
answered Feb 29 '12 at 6:26
Brian M. ScottBrian M. Scott
461k40518920
$endgroup$
add a comment |
$begingroup$
Henno’s argument is probably the slickest, but one can also work pointwise, either directly or, if one already has those characterizations of continuity, with nets or filters.
Suppose that $f$ is not continuous. Then there is some point $x_0in X$ at which $f$ fails to be continuous. If $y_0=f(x_0)$, this means that there is an open nbhd $U$ of $y_0$ such that if $V$ is an open nbhd of $x_0$, there is a point $x_Vin V$ such that $f(x_V)notin U$. Let $$A={x_V:Vtext{ is an open nbhd of }x_0};.$$
Clearly $x_0inoverline{A}$, and $f[A]subseteq Ysetminus U$. Moreover, $Ysetminus U$ is closed, so $overline{f[A]}subseteq Ysetminus U$, and hence
$$y_0in f[overline{A}]subseteqoverline{f[A]}subseteq Ysetminus U;,$$
contradicting the choice of $U$.
Those who like nets may notice that $A$ actually is a net, over the directed set $langlemathscr{N}(x_0),supseteqrangle$, where $mathscr{N}(x_0)$ is the family of open nbhds of $x_0$, and the argument shows that $f$ doesn’t preserve the limit of this net. (I could of course just as well have used the nbhd filter at $x_0$, instead of using only open nbhds.)
Those who prefer filters may modify this to consider the filter $$mathscr{F}={Vcap f^{-1}[Ysetminus U]:Vinmathscr{N}(x_0)}$$ instead.
answered Feb 29 '12 at 6:26
Brian M. ScottBrian M. Scott
461k40518920
$endgroup$
Henno’s argument is probably the slickest, but one can also work pointwise, either directly or, if one already has those characterizations of continuity, with nets or filters.
Suppose that $f$ is not continuous. Then there is some point $x_0in X$ at which $f$ fails to be continuous. If $y_0=f(x_0)$, this means that there is an open nbhd $U$ of $y_0$ such that if $V$ is an open nbhd of $x_0$, there is a point $x_Vin V$ such that $f(x_V)notin U$. Let $$A={x_V:Vtext{ is an open nbhd of }x_0};.$$
Clearly $x_0inoverline{A}$, and $f[A]subseteq Ysetminus U$. Moreover, $Ysetminus U$ is closed, so $overline{f[A]}subseteq Ysetminus U$, and hence
$$y_0in f[overline{A}]subseteqoverline{f[A]}subseteq Ysetminus U;,$$
contradicting the choice of $U$.
Those who like nets may notice that $A$ actually is a net, over the directed set $langlemathscr{N}(x_0),supseteqrangle$, where $mathscr{N}(x_0)$ is the family of open nbhds of $x_0$, and the argument shows that $f$ doesn’t preserve the limit of this net. (I could of course just as well have used the nbhd filter at $x_0$, instead of using only open nbhds.)
Those who prefer filters may modify this to consider the filter $$mathscr{F}={Vcap f^{-1}[Ysetminus U]:Vinmathscr{N}(x_0)}$$ instead.
answered Feb 29 '12 at 6:26
Brian M. ScottBrian M. Scott
461k40518920
answered Feb 29 '12 at 6:26
Brian M. ScottBrian M. Scott
461k40518920
answered Feb 29 '12 at 6:26
Brian M. ScottBrian M. Scott
461k40518920
answered Feb 29 '12 at 6:26
Brian M. ScottBrian M. Scott
461k40518920
461k40518920
add a comment |
add a comment |
$begingroup$
Here's one proof of the converse provided $X$ and $Y$ are metric spaces:
Take a limit point $x$ of $A$. Then because $f(overline{A}) subseteq overline{f(A)}$, we have that $f(x)$ is a limit point of $f(A)$.
Because $x$ is a limit point of $A$, for every $delta > 0$ there is a point $p in A$ with $|x - p| < delta$. And because $f(x)$ is a limit point of $f(A)$, for every $epsilon > 0$, there is a point $q in f(A)$ with $|f(x) - f(p)| < epsilon$.
We thus have that for every point $x in A$ with $|x - p| < delta$, in fact $|f(x) - f(p)| < epsilon$, so $f$ must be continuous.
edited Feb 28 '12 at 17:11
Austin Mohr
20.9k35299
answered Feb 28 '12 at 16:33
jamaicanwormjamaicanworm
1,17762638
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4
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I guess this is a decent enough answer for the case of the real numbers, but the question was about arbitrary topological spaces $X$ and $Y$.
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– kahen
Feb 28 '12 at 16:45
3
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We are not in a world where things like $|x-p|$ could make sense. Any "TOPOLOGICAL SPACE" is in question, but any way a good proof that will be useful. I am upvoting and don't delete this answer because of this as it will help close duplicates.
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– user21436
Feb 28 '12 at 16:46
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Can't you just substitute $d(x, p)$ for $|x - p|$?
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– jamaicanworm
Feb 28 '12 at 16:50
4
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@jamaicaworm: Only in a metric space. Not every topological space is a metric space, or even metrizable.
$endgroup$
– Arturo Magidin
Feb 28 '12 at 16:50
6
$begingroup$
@jamaicanworm: The post is labeled [general-topology], not [metric-spaces]. The definition of a topological space includes the notion of "open set", and "closed set" is a set whose complement is open. See the definition in Wikipedia.
$endgroup$
– Arturo Magidin
Feb 28 '12 at 16:59
|
show 1 more comment
$begingroup$
Here's one proof of the converse provided $X$ and $Y$ are metric spaces:
Take a limit point $x$ of $A$. Then because $f(overline{A}) subseteq overline{f(A)}$, we have that $f(x)$ is a limit point of $f(A)$.
Because $x$ is a limit point of $A$, for every $delta > 0$ there is a point $p in A$ with $|x - p| < delta$. And because $f(x)$ is a limit point of $f(A)$, for every $epsilon > 0$, there is a point $q in f(A)$ with $|f(x) - f(p)| < epsilon$.
We thus have that for every point $x in A$ with $|x - p| < delta$, in fact $|f(x) - f(p)| < epsilon$, so $f$ must be continuous.
edited Feb 28 '12 at 17:11
Austin Mohr
20.9k35299
answered Feb 28 '12 at 16:33
jamaicanwormjamaicanworm
1,17762638
$endgroup$
4
$begingroup$
I guess this is a decent enough answer for the case of the real numbers, but the question was about arbitrary topological spaces $X$ and $Y$.
$endgroup$
– kahen
Feb 28 '12 at 16:45
3
$begingroup$
We are not in a world where things like $|x-p|$ could make sense. Any "TOPOLOGICAL SPACE" is in question, but any way a good proof that will be useful. I am upvoting and don't delete this answer because of this as it will help close duplicates.
$endgroup$
– user21436
Feb 28 '12 at 16:46
$begingroup$
Can't you just substitute $d(x, p)$ for $|x - p|$?
$endgroup$
– jamaicanworm
Feb 28 '12 at 16:50
4
$begingroup$
@jamaicaworm: Only in a metric space. Not every topological space is a metric space, or even metrizable.
$endgroup$
– Arturo Magidin
Feb 28 '12 at 16:50
6
$begingroup$
@jamaicanworm: The post is labeled [general-topology], not [metric-spaces]. The definition of a topological space includes the notion of "open set", and "closed set" is a set whose complement is open. See the definition in Wikipedia.
$endgroup$
– Arturo Magidin
Feb 28 '12 at 16:59
|
show 1 more comment
$begingroup$
Here's one proof of the converse provided $X$ and $Y$ are metric spaces:
Take a limit point $x$ of $A$. Then because $f(overline{A}) subseteq overline{f(A)}$, we have that $f(x)$ is a limit point of $f(A)$.
Because $x$ is a limit point of $A$, for every $delta > 0$ there is a point $p in A$ with $|x - p| < delta$. And because $f(x)$ is a limit point of $f(A)$, for every $epsilon > 0$, there is a point $q in f(A)$ with $|f(x) - f(p)| < epsilon$.
We thus have that for every point $x in A$ with $|x - p| < delta$, in fact $|f(x) - f(p)| < epsilon$, so $f$ must be continuous.
edited Feb 28 '12 at 17:11
Austin Mohr
20.9k35299
answered Feb 28 '12 at 16:33
jamaicanwormjamaicanworm
1,17762638
$endgroup$
Here's one proof of the converse provided $X$ and $Y$ are metric spaces:
Take a limit point $x$ of $A$. Then because $f(overline{A}) subseteq overline{f(A)}$, we have that $f(x)$ is a limit point of $f(A)$.
Because $x$ is a limit point of $A$, for every $delta > 0$ there is a point $p in A$ with $|x - p| < delta$. And because $f(x)$ is a limit point of $f(A)$, for every $epsilon > 0$, there is a point $q in f(A)$ with $|f(x) - f(p)| < epsilon$.
We thus have that for every point $x in A$ with $|x - p| < delta$, in fact $|f(x) - f(p)| < epsilon$, so $f$ must be continuous.
edited Feb 28 '12 at 17:11
Austin Mohr
20.9k35299
answered Feb 28 '12 at 16:33
jamaicanwormjamaicanworm
1,17762638
edited Feb 28 '12 at 17:11
Austin Mohr
20.9k35299
edited Feb 28 '12 at 17:11
Austin Mohr
20.9k35299
edited Feb 28 '12 at 17:11
Austin Mohr
20.9k35299
20.9k35299
answered Feb 28 '12 at 16:33
jamaicanwormjamaicanworm
1,17762638
answered Feb 28 '12 at 16:33
jamaicanwormjamaicanworm
1,17762638
answered Feb 28 '12 at 16:33
jamaicanwormjamaicanworm
1,17762638
1,17762638
4
$begingroup$
I guess this is a decent enough answer for the case of the real numbers, but the question was about arbitrary topological spaces $X$ and $Y$.
$endgroup$
– kahen
Feb 28 '12 at 16:45
3
$begingroup$
We are not in a world where things like $|x-p|$ could make sense. Any "TOPOLOGICAL SPACE" is in question, but any way a good proof that will be useful. I am upvoting and don't delete this answer because of this as it will help close duplicates.
$endgroup$
– user21436
Feb 28 '12 at 16:46
$begingroup$
Can't you just substitute $d(x, p)$ for $|x - p|$?
$endgroup$
– jamaicanworm
Feb 28 '12 at 16:50
4
$begingroup$
@jamaicaworm: Only in a metric space. Not every topological space is a metric space, or even metrizable.
$endgroup$
– Arturo Magidin
Feb 28 '12 at 16:50
6
$begingroup$
@jamaicanworm: The post is labeled [general-topology], not [metric-spaces]. The definition of a topological space includes the notion of "open set", and "closed set" is a set whose complement is open. See the definition in Wikipedia.
$endgroup$
– Arturo Magidin
Feb 28 '12 at 16:59
|
show 1 more comment
4
$begingroup$
I guess this is a decent enough answer for the case of the real numbers, but the question was about arbitrary topological spaces $X$ and $Y$.
$endgroup$
– kahen
Feb 28 '12 at 16:45
3
$begingroup$
We are not in a world where things like $|x-p|$ could make sense. Any "TOPOLOGICAL SPACE" is in question, but any way a good proof that will be useful. I am upvoting and don't delete this answer because of this as it will help close duplicates.
$endgroup$
– user21436
Feb 28 '12 at 16:46
$begingroup$
Can't you just substitute $d(x, p)$ for $|x - p|$?
$endgroup$
– jamaicanworm
Feb 28 '12 at 16:50
4
$begingroup$
@jamaicaworm: Only in a metric space. Not every topological space is a metric space, or even metrizable.
$endgroup$
– Arturo Magidin
Feb 28 '12 at 16:50
6
$begingroup$
@jamaicanworm: The post is labeled [general-topology], not [metric-spaces]. The definition of a topological space includes the notion of "open set", and "closed set" is a set whose complement is open. See the definition in Wikipedia.
$endgroup$
– Arturo Magidin
Feb 28 '12 at 16:59
4
4
$begingroup$
I guess this is a decent enough answer for the case of the real numbers, but the question was about arbitrary topological spaces $X$ and $Y$.
$endgroup$
– kahen
Feb 28 '12 at 16:45
$begingroup$
I guess this is a decent enough answer for the case of the real numbers, but the question was about arbitrary topological spaces $X$ and $Y$.
$endgroup$
– kahen
Feb 28 '12 at 16:45
3
3
$begingroup$
We are not in a world where things like $|x-p|$ could make sense. Any "TOPOLOGICAL SPACE" is in question, but any way a good proof that will be useful. I am upvoting and don't delete this answer because of this as it will help close duplicates.
$endgroup$
– user21436
Feb 28 '12 at 16:46
$begingroup$
We are not in a world where things like $|x-p|$ could make sense. Any "TOPOLOGICAL SPACE" is in question, but any way a good proof that will be useful. I am upvoting and don't delete this answer because of this as it will help close duplicates.
$endgroup$
– user21436
Feb 28 '12 at 16:46
$begingroup$
Can't you just substitute $d(x, p)$ for $|x - p|$?
$endgroup$
– jamaicanworm
Feb 28 '12 at 16:50
$begingroup$
Can't you just substitute $d(x, p)$ for $|x - p|$?
$endgroup$
– jamaicanworm
Feb 28 '12 at 16:50
4
4
$begingroup$
@jamaicaworm: Only in a metric space. Not every topological space is a metric space, or even metrizable.
$endgroup$
– Arturo Magidin
Feb 28 '12 at 16:50
$begingroup$
@jamaicaworm: Only in a metric space. Not every topological space is a metric space, or even metrizable.
$endgroup$
– Arturo Magidin
Feb 28 '12 at 16:50
6
6
$begingroup$
@jamaicanworm: The post is labeled [general-topology], not [metric-spaces]. The definition of a topological space includes the notion of "open set", and "closed set" is a set whose complement is open. See the definition in Wikipedia.
$endgroup$
– Arturo Magidin
Feb 28 '12 at 16:59
$begingroup$
@jamaicanworm: The post is labeled [general-topology], not [metric-spaces]. The definition of a topological space includes the notion of "open set", and "closed set" is a set whose complement is open. See the definition in Wikipedia.
$endgroup$
– Arturo Magidin
Feb 28 '12 at 16:59
|
show 1 more comment
$begingroup$
A proof using nets:
Suppose $yin f(overline{A})backslash overline{fA}$. Taking $xin overline{A}$ with $y=f(x)$; there is a net $x_iin A$ converging to $x$. Now $f(x_i)$ is a net in $f(A)$ which does not converge to $f(x)$ [for this would imply $f(x)in overline{fA}$]. Thus $f$ is not continuous.
answered Feb 28 '12 at 22:02
wildildildlifewildildildlife
4,5142222
$endgroup$
$begingroup$
This is the direction that the OP had already been able to show.
$endgroup$
– Brian M. Scott
Feb 29 '12 at 6:08
$begingroup$
@Brian: Yes, sorry! I realized this the day after...
$endgroup$
– wildildildlife
Mar 2 '12 at 13:15
add a comment |
$begingroup$
A proof using nets:
Suppose $yin f(overline{A})backslash overline{fA}$. Taking $xin overline{A}$ with $y=f(x)$; there is a net $x_iin A$ converging to $x$. Now $f(x_i)$ is a net in $f(A)$ which does not converge to $f(x)$ [for this would imply $f(x)in overline{fA}$]. Thus $f$ is not continuous.
answered Feb 28 '12 at 22:02
wildildildlifewildildildlife
4,5142222
$endgroup$
$begingroup$
This is the direction that the OP had already been able to show.
$endgroup$
– Brian M. Scott
Feb 29 '12 at 6:08
$begingroup$
@Brian: Yes, sorry! I realized this the day after...
$endgroup$
– wildildildlife
Mar 2 '12 at 13:15
add a comment |
$begingroup$
A proof using nets:
Suppose $yin f(overline{A})backslash overline{fA}$. Taking $xin overline{A}$ with $y=f(x)$; there is a net $x_iin A$ converging to $x$. Now $f(x_i)$ is a net in $f(A)$ which does not converge to $f(x)$ [for this would imply $f(x)in overline{fA}$]. Thus $f$ is not continuous.
answered Feb 28 '12 at 22:02
wildildildlifewildildildlife
4,5142222
$endgroup$
A proof using nets:
Suppose $yin f(overline{A})backslash overline{fA}$. Taking $xin overline{A}$ with $y=f(x)$; there is a net $x_iin A$ converging to $x$. Now $f(x_i)$ is a net in $f(A)$ which does not converge to $f(x)$ [for this would imply $f(x)in overline{fA}$]. Thus $f$ is not continuous.
answered Feb 28 '12 at 22:02
wildildildlifewildildildlife
4,5142222
answered Feb 28 '12 at 22:02
wildildildlifewildildildlife
4,5142222
answered Feb 28 '12 at 22:02
wildildildlifewildildildlife
4,5142222
answered Feb 28 '12 at 22:02
wildildildlifewildildildlife
4,5142222
4,5142222
$begingroup$
This is the direction that the OP had already been able to show.
$endgroup$
– Brian M. Scott
Feb 29 '12 at 6:08
$begingroup$
@Brian: Yes, sorry! I realized this the day after...
$endgroup$
– wildildildlife
Mar 2 '12 at 13:15
add a comment |
$begingroup$
This is the direction that the OP had already been able to show.
$endgroup$
– Brian M. Scott
Feb 29 '12 at 6:08
$begingroup$
@Brian: Yes, sorry! I realized this the day after...
$endgroup$
– wildildildlife
Mar 2 '12 at 13:15
$begingroup$
This is the direction that the OP had already been able to show.
$endgroup$
– Brian M. Scott
Feb 29 '12 at 6:08
$begingroup$
This is the direction that the OP had already been able to show.
$endgroup$
– Brian M. Scott
Feb 29 '12 at 6:08
$begingroup$
@Brian: Yes, sorry! I realized this the day after...
$endgroup$
– wildildildlife
Mar 2 '12 at 13:15
$begingroup$
@Brian: Yes, sorry! I realized this the day after...
$endgroup$
– wildildildlife
Mar 2 '12 at 13:15
add a comment |
$begingroup$
The assertion is equivalent to:
$overline{A}subseteq f^{-1}(overline{f(A)})$
So, the assertion follows from:
$overline{A}subseteqoverline{f^{-1}(f(A))}subseteqoverline{f^{-1}(overline{f(A)})}=f^{-1}(overline{f(A)})$
- Inclusion: $Asubseteq f^{-1}(f(A)) Rightarrow overline{A}subseteqoverline{f^{-1}(f(A))}$
- Inclusion: $f(A)subseteqoverline{f(A)} Rightarrow f^{-1}(f(A))subseteq f^{-1}(overline{f(A)}) Rightarrow overline{f^{-1}(f(A))}subseteq overline{f^{-1}(overline{f(A)})}$
- Equality: $overline{f(A)} text{ closed} Rightarrow f^{-1}(overline{f(A)}) text{ closed} Rightarrow overline{f^{-1}(overline{f(A)})}=f^{-1}(overline{f(A)})$
The converse assertion is equivalent to:
$overline{B}=B Rightarrow overline{f^{-1}(B)}=f^{-1}(B)$
So, the converse assertion follows from:
$f^{-1}(B)subseteqoverline{f^{-1}(B)}subseteq f^{-1}(f(overline{f^{-1}(B)}))subseteq f^{-1}(overline{f(f^{-1}(B))}) subseteq f^{-1}(overline{B}) =f^{-1}(B)$
That gives:
$f^{-1}(B)=overline{f^{-1}(B)}$
- Inclusion: $Asubseteq overline{A} text{ in general}$
- Inclusion: $Asubseteq f^{-1}(f(A)) text{ in general}$
- Inclusion: $f(overline{A})subseteq overline{f(A)} text{ by assumption}$
- Inclusion: $f(f^{-1}(B))subseteq B text{ in general} Rightarrow overline{f(f^{-1}(B))}subseteq overline{B} Rightarrow f^{-1}(overline{f(f^{-1}(B))})subseteq f^{-1}(overline{B})$
- Equality: $overline{B}=B Rightarrow f^{-1}(overline{B})=f^{-1}(B)$
answered Jan 22 '14 at 17:38
C-Star-W-StarC-Star-W-Star
8,36642166
$endgroup$
1
$begingroup$
I don't see how this is shorter than Henno's answer, which is complete since the OP cleared out the other direction. I don't think it is good practice to claim your answer is somehow better than others, and specifically target another user's answer.
$endgroup$
– Pedro Tamaroff♦
Jan 22 '14 at 17:51
$begingroup$
I'm sorry for that ...got a little bit upset... I deleted that part so there's no bad discussion about it -sorry for that! Anyway Henno only shows that the preimage of a closed sst is closed what is a rather simple consequence. But the assertion is much more tricky than that. See my comment on Henno's answer.
$endgroup$
– C-Star-W-Star
Jan 22 '14 at 17:57
add a comment |
$begingroup$
The assertion is equivalent to:
$overline{A}subseteq f^{-1}(overline{f(A)})$
So, the assertion follows from:
$overline{A}subseteqoverline{f^{-1}(f(A))}subseteqoverline{f^{-1}(overline{f(A)})}=f^{-1}(overline{f(A)})$
- Inclusion: $Asubseteq f^{-1}(f(A)) Rightarrow overline{A}subseteqoverline{f^{-1}(f(A))}$
- Inclusion: $f(A)subseteqoverline{f(A)} Rightarrow f^{-1}(f(A))subseteq f^{-1}(overline{f(A)}) Rightarrow overline{f^{-1}(f(A))}subseteq overline{f^{-1}(overline{f(A)})}$
- Equality: $overline{f(A)} text{ closed} Rightarrow f^{-1}(overline{f(A)}) text{ closed} Rightarrow overline{f^{-1}(overline{f(A)})}=f^{-1}(overline{f(A)})$
The converse assertion is equivalent to:
$overline{B}=B Rightarrow overline{f^{-1}(B)}=f^{-1}(B)$
So, the converse assertion follows from:
$f^{-1}(B)subseteqoverline{f^{-1}(B)}subseteq f^{-1}(f(overline{f^{-1}(B)}))subseteq f^{-1}(overline{f(f^{-1}(B))}) subseteq f^{-1}(overline{B}) =f^{-1}(B)$
That gives:
$f^{-1}(B)=overline{f^{-1}(B)}$
- Inclusion: $Asubseteq overline{A} text{ in general}$
- Inclusion: $Asubseteq f^{-1}(f(A)) text{ in general}$
- Inclusion: $f(overline{A})subseteq overline{f(A)} text{ by assumption}$
- Inclusion: $f(f^{-1}(B))subseteq B text{ in general} Rightarrow overline{f(f^{-1}(B))}subseteq overline{B} Rightarrow f^{-1}(overline{f(f^{-1}(B))})subseteq f^{-1}(overline{B})$
- Equality: $overline{B}=B Rightarrow f^{-1}(overline{B})=f^{-1}(B)$
answered Jan 22 '14 at 17:38
C-Star-W-StarC-Star-W-Star
8,36642166
$endgroup$
1
$begingroup$
I don't see how this is shorter than Henno's answer, which is complete since the OP cleared out the other direction. I don't think it is good practice to claim your answer is somehow better than others, and specifically target another user's answer.
$endgroup$
– Pedro Tamaroff♦
Jan 22 '14 at 17:51
$begingroup$
I'm sorry for that ...got a little bit upset... I deleted that part so there's no bad discussion about it -sorry for that! Anyway Henno only shows that the preimage of a closed sst is closed what is a rather simple consequence. But the assertion is much more tricky than that. See my comment on Henno's answer.
$endgroup$
– C-Star-W-Star
Jan 22 '14 at 17:57
add a comment |
$begingroup$
The assertion is equivalent to:
$overline{A}subseteq f^{-1}(overline{f(A)})$
So, the assertion follows from:
$overline{A}subseteqoverline{f^{-1}(f(A))}subseteqoverline{f^{-1}(overline{f(A)})}=f^{-1}(overline{f(A)})$
- Inclusion: $Asubseteq f^{-1}(f(A)) Rightarrow overline{A}subseteqoverline{f^{-1}(f(A))}$
- Inclusion: $f(A)subseteqoverline{f(A)} Rightarrow f^{-1}(f(A))subseteq f^{-1}(overline{f(A)}) Rightarrow overline{f^{-1}(f(A))}subseteq overline{f^{-1}(overline{f(A)})}$
- Equality: $overline{f(A)} text{ closed} Rightarrow f^{-1}(overline{f(A)}) text{ closed} Rightarrow overline{f^{-1}(overline{f(A)})}=f^{-1}(overline{f(A)})$
The converse assertion is equivalent to:
$overline{B}=B Rightarrow overline{f^{-1}(B)}=f^{-1}(B)$
So, the converse assertion follows from:
$f^{-1}(B)subseteqoverline{f^{-1}(B)}subseteq f^{-1}(f(overline{f^{-1}(B)}))subseteq f^{-1}(overline{f(f^{-1}(B))}) subseteq f^{-1}(overline{B}) =f^{-1}(B)$
That gives:
$f^{-1}(B)=overline{f^{-1}(B)}$
- Inclusion: $Asubseteq overline{A} text{ in general}$
- Inclusion: $Asubseteq f^{-1}(f(A)) text{ in general}$
- Inclusion: $f(overline{A})subseteq overline{f(A)} text{ by assumption}$
- Inclusion: $f(f^{-1}(B))subseteq B text{ in general} Rightarrow overline{f(f^{-1}(B))}subseteq overline{B} Rightarrow f^{-1}(overline{f(f^{-1}(B))})subseteq f^{-1}(overline{B})$
- Equality: $overline{B}=B Rightarrow f^{-1}(overline{B})=f^{-1}(B)$
answered Jan 22 '14 at 17:38
C-Star-W-StarC-Star-W-Star
8,36642166
$endgroup$
The assertion is equivalent to:
$overline{A}subseteq f^{-1}(overline{f(A)})$
So, the assertion follows from:
$overline{A}subseteqoverline{f^{-1}(f(A))}subseteqoverline{f^{-1}(overline{f(A)})}=f^{-1}(overline{f(A)})$
- Inclusion: $Asubseteq f^{-1}(f(A)) Rightarrow overline{A}subseteqoverline{f^{-1}(f(A))}$
- Inclusion: $f(A)subseteqoverline{f(A)} Rightarrow f^{-1}(f(A))subseteq f^{-1}(overline{f(A)}) Rightarrow overline{f^{-1}(f(A))}subseteq overline{f^{-1}(overline{f(A)})}$
- Equality: $overline{f(A)} text{ closed} Rightarrow f^{-1}(overline{f(A)}) text{ closed} Rightarrow overline{f^{-1}(overline{f(A)})}=f^{-1}(overline{f(A)})$
The converse assertion is equivalent to:
$overline{B}=B Rightarrow overline{f^{-1}(B)}=f^{-1}(B)$
So, the converse assertion follows from:
$f^{-1}(B)subseteqoverline{f^{-1}(B)}subseteq f^{-1}(f(overline{f^{-1}(B)}))subseteq f^{-1}(overline{f(f^{-1}(B))}) subseteq f^{-1}(overline{B}) =f^{-1}(B)$
That gives:
$f^{-1}(B)=overline{f^{-1}(B)}$
- Inclusion: $Asubseteq overline{A} text{ in general}$
- Inclusion: $Asubseteq f^{-1}(f(A)) text{ in general}$
- Inclusion: $f(overline{A})subseteq overline{f(A)} text{ by assumption}$
- Inclusion: $f(f^{-1}(B))subseteq B text{ in general} Rightarrow overline{f(f^{-1}(B))}subseteq overline{B} Rightarrow f^{-1}(overline{f(f^{-1}(B))})subseteq f^{-1}(overline{B})$
- Equality: $overline{B}=B Rightarrow f^{-1}(overline{B})=f^{-1}(B)$
answered Jan 22 '14 at 17:38
C-Star-W-StarC-Star-W-Star
8,36642166
edited Jan 22 '14 at 18:35
answered Jan 22 '14 at 17:38
C-Star-W-StarC-Star-W-Star
8,36642166
answered Jan 22 '14 at 17:38
C-Star-W-StarC-Star-W-Star
8,36642166
answered Jan 22 '14 at 17:38
C-Star-W-StarC-Star-W-Star
8,36642166
8,36642166
1
$begingroup$
I don't see how this is shorter than Henno's answer, which is complete since the OP cleared out the other direction. I don't think it is good practice to claim your answer is somehow better than others, and specifically target another user's answer.
$endgroup$
– Pedro Tamaroff♦
Jan 22 '14 at 17:51
$begingroup$
I'm sorry for that ...got a little bit upset... I deleted that part so there's no bad discussion about it -sorry for that! Anyway Henno only shows that the preimage of a closed sst is closed what is a rather simple consequence. But the assertion is much more tricky than that. See my comment on Henno's answer.
$endgroup$
– C-Star-W-Star
Jan 22 '14 at 17:57
add a comment |
1
$begingroup$
I don't see how this is shorter than Henno's answer, which is complete since the OP cleared out the other direction. I don't think it is good practice to claim your answer is somehow better than others, and specifically target another user's answer.
$endgroup$
– Pedro Tamaroff♦
Jan 22 '14 at 17:51
$begingroup$
I'm sorry for that ...got a little bit upset... I deleted that part so there's no bad discussion about it -sorry for that! Anyway Henno only shows that the preimage of a closed sst is closed what is a rather simple consequence. But the assertion is much more tricky than that. See my comment on Henno's answer.
$endgroup$
– C-Star-W-Star
Jan 22 '14 at 17:57
1
1
$begingroup$
I don't see how this is shorter than Henno's answer, which is complete since the OP cleared out the other direction. I don't think it is good practice to claim your answer is somehow better than others, and specifically target another user's answer.
$endgroup$
– Pedro Tamaroff♦
Jan 22 '14 at 17:51
$begingroup$
I don't see how this is shorter than Henno's answer, which is complete since the OP cleared out the other direction. I don't think it is good practice to claim your answer is somehow better than others, and specifically target another user's answer.
$endgroup$
– Pedro Tamaroff♦
Jan 22 '14 at 17:51
$begingroup$
I'm sorry for that ...got a little bit upset... I deleted that part so there's no bad discussion about it -sorry for that! Anyway Henno only shows that the preimage of a closed sst is closed what is a rather simple consequence. But the assertion is much more tricky than that. See my comment on Henno's answer.
$endgroup$
– C-Star-W-Star
Jan 22 '14 at 17:57
$begingroup$
I'm sorry for that ...got a little bit upset... I deleted that part so there's no bad discussion about it -sorry for that! Anyway Henno only shows that the preimage of a closed sst is closed what is a rather simple consequence. But the assertion is much more tricky than that. See my comment on Henno's answer.
$endgroup$
– C-Star-W-Star
Jan 22 '14 at 17:57
add a comment |
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$begingroup$
I am not able to come up with any example of a continuous function s.t. $f(overline{A}) subsetneq overline{f(A)}$. Can anyone give such an examples?
$endgroup$
– MUH
Mar 11 '18 at 11:03
2
$begingroup$
Assuming $f$ is continuous, how exactly is the result "immediate"?
$endgroup$
– Al Jebr
May 19 '18 at 18:56