If $f: mathbb{R} to mathbb{R}$ is a continuous surjection, must it be open?
$begingroup$
If $f: mathbb{R} to mathbb{R}$ is a continuous surjection, must it be open?
I think not. I proved if $f: mathbb{R} to mathbb{R}$ is an open continuous surjection, then $f$ is a homeomorphism. So, if the question is true, every continuous surjection must be a homeomorphism. But, I didn't find a counterexample. Can someone help me?
real-analysis general-topology metric-spaces
asked Dec 20 '18 at 18:39
Lucas CorrêaLucas Corrêa
1,5471421
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add a comment |
$begingroup$
If $f: mathbb{R} to mathbb{R}$ is a continuous surjection, must it be open?
I think not. I proved if $f: mathbb{R} to mathbb{R}$ is an open continuous surjection, then $f$ is a homeomorphism. So, if the question is true, every continuous surjection must be a homeomorphism. But, I didn't find a counterexample. Can someone help me?
real-analysis general-topology metric-spaces
asked Dec 20 '18 at 18:39
Lucas CorrêaLucas Corrêa
1,5471421
$endgroup$
2
$begingroup$
What about a continuous surjections that is constant on an interval? for example,$f(x)=x$ for $xle 0$, $f(x)=0$ for $0<x<1$ and $f(x)=x-1$ for $xge 1$.
$endgroup$
– Tito Eliatron
Dec 20 '18 at 18:42
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This is very helpful! Thanks!
$endgroup$
– Lucas Corrêa
Dec 20 '18 at 18:46
2
$begingroup$
If you want one with a closed formula, you can take $f(x)=x^3-x$. Plotting this should make the properties clear.
$endgroup$
– SmileyCraft
Dec 20 '18 at 18:47
add a comment |
$begingroup$
If $f: mathbb{R} to mathbb{R}$ is a continuous surjection, must it be open?
I think not. I proved if $f: mathbb{R} to mathbb{R}$ is an open continuous surjection, then $f$ is a homeomorphism. So, if the question is true, every continuous surjection must be a homeomorphism. But, I didn't find a counterexample. Can someone help me?
real-analysis general-topology metric-spaces
asked Dec 20 '18 at 18:39
Lucas CorrêaLucas Corrêa
1,5471421
$endgroup$
If $f: mathbb{R} to mathbb{R}$ is a continuous surjection, must it be open?
I think not. I proved if $f: mathbb{R} to mathbb{R}$ is an open continuous surjection, then $f$ is a homeomorphism. So, if the question is true, every continuous surjection must be a homeomorphism. But, I didn't find a counterexample. Can someone help me?
real-analysis general-topology metric-spaces
real-analysis general-topology metric-spaces
asked Dec 20 '18 at 18:39
Lucas CorrêaLucas Corrêa
1,5471421
asked Dec 20 '18 at 18:39
Lucas CorrêaLucas Corrêa
1,5471421
asked Dec 20 '18 at 18:39
Lucas CorrêaLucas Corrêa
1,5471421
asked Dec 20 '18 at 18:39
Lucas CorrêaLucas Corrêa
1,5471421
asked Dec 20 '18 at 18:39
Lucas CorrêaLucas Corrêa
1,5471421
1,5471421
2
$begingroup$
What about a continuous surjections that is constant on an interval? for example,$f(x)=x$ for $xle 0$, $f(x)=0$ for $0<x<1$ and $f(x)=x-1$ for $xge 1$.
$endgroup$
– Tito Eliatron
Dec 20 '18 at 18:42
$begingroup$
This is very helpful! Thanks!
$endgroup$
– Lucas Corrêa
Dec 20 '18 at 18:46
2
$begingroup$
If you want one with a closed formula, you can take $f(x)=x^3-x$. Plotting this should make the properties clear.
$endgroup$
– SmileyCraft
Dec 20 '18 at 18:47
add a comment |
2
$begingroup$
What about a continuous surjections that is constant on an interval? for example,$f(x)=x$ for $xle 0$, $f(x)=0$ for $0<x<1$ and $f(x)=x-1$ for $xge 1$.
$endgroup$
– Tito Eliatron
Dec 20 '18 at 18:42
$begingroup$
This is very helpful! Thanks!
$endgroup$
– Lucas Corrêa
Dec 20 '18 at 18:46
2
$begingroup$
If you want one with a closed formula, you can take $f(x)=x^3-x$. Plotting this should make the properties clear.
$endgroup$
– SmileyCraft
Dec 20 '18 at 18:47
2
2
$begingroup$
What about a continuous surjections that is constant on an interval? for example,$f(x)=x$ for $xle 0$, $f(x)=0$ for $0<x<1$ and $f(x)=x-1$ for $xge 1$.
$endgroup$
– Tito Eliatron
Dec 20 '18 at 18:42
$begingroup$
What about a continuous surjections that is constant on an interval? for example,$f(x)=x$ for $xle 0$, $f(x)=0$ for $0<x<1$ and $f(x)=x-1$ for $xge 1$.
$endgroup$
– Tito Eliatron
Dec 20 '18 at 18:42
$begingroup$
This is very helpful! Thanks!
$endgroup$
– Lucas Corrêa
Dec 20 '18 at 18:46
$begingroup$
This is very helpful! Thanks!
$endgroup$
– Lucas Corrêa
Dec 20 '18 at 18:46
2
2
$begingroup$
If you want one with a closed formula, you can take $f(x)=x^3-x$. Plotting this should make the properties clear.
$endgroup$
– SmileyCraft
Dec 20 '18 at 18:47
$begingroup$
If you want one with a closed formula, you can take $f(x)=x^3-x$. Plotting this should make the properties clear.
$endgroup$
– SmileyCraft
Dec 20 '18 at 18:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Any surjection that attains a local extremum suffices. Consider for instance
$$
f(x) = x(x-1)(x-2) = x^3 - 3x^2 + 3x
$$
$f$ is not an open map since the interval $(0,1)$ is mapped to an interval of the form $(0,a]$, which is not open.
answered Dec 20 '18 at 18:48
OmnomnomnomOmnomnomnom
129k792185
$endgroup$
1
$begingroup$
Looks like SmileyCraft had the same idea
$endgroup$
– Omnomnomnom
Dec 20 '18 at 18:49
add a comment |
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1 Answer
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$begingroup$
Any surjection that attains a local extremum suffices. Consider for instance
$$
f(x) = x(x-1)(x-2) = x^3 - 3x^2 + 3x
$$
$f$ is not an open map since the interval $(0,1)$ is mapped to an interval of the form $(0,a]$, which is not open.
answered Dec 20 '18 at 18:48
OmnomnomnomOmnomnomnom
129k792185
$endgroup$
1
$begingroup$
Looks like SmileyCraft had the same idea
$endgroup$
– Omnomnomnom
Dec 20 '18 at 18:49
add a comment |
$begingroup$
Any surjection that attains a local extremum suffices. Consider for instance
$$
f(x) = x(x-1)(x-2) = x^3 - 3x^2 + 3x
$$
$f$ is not an open map since the interval $(0,1)$ is mapped to an interval of the form $(0,a]$, which is not open.
answered Dec 20 '18 at 18:48
OmnomnomnomOmnomnomnom
129k792185
$endgroup$
1
$begingroup$
Looks like SmileyCraft had the same idea
$endgroup$
– Omnomnomnom
Dec 20 '18 at 18:49
add a comment |
$begingroup$
Any surjection that attains a local extremum suffices. Consider for instance
$$
f(x) = x(x-1)(x-2) = x^3 - 3x^2 + 3x
$$
$f$ is not an open map since the interval $(0,1)$ is mapped to an interval of the form $(0,a]$, which is not open.
answered Dec 20 '18 at 18:48
OmnomnomnomOmnomnomnom
129k792185
$endgroup$
Any surjection that attains a local extremum suffices. Consider for instance
$$
f(x) = x(x-1)(x-2) = x^3 - 3x^2 + 3x
$$
$f$ is not an open map since the interval $(0,1)$ is mapped to an interval of the form $(0,a]$, which is not open.
answered Dec 20 '18 at 18:48
OmnomnomnomOmnomnomnom
129k792185
answered Dec 20 '18 at 18:48
OmnomnomnomOmnomnomnom
129k792185
answered Dec 20 '18 at 18:48
OmnomnomnomOmnomnomnom
129k792185
answered Dec 20 '18 at 18:48
OmnomnomnomOmnomnomnom
129k792185
129k792185
1
$begingroup$
Looks like SmileyCraft had the same idea
$endgroup$
– Omnomnomnom
Dec 20 '18 at 18:49
add a comment |
1
$begingroup$
Looks like SmileyCraft had the same idea
$endgroup$
– Omnomnomnom
Dec 20 '18 at 18:49
1
1
$begingroup$
Looks like SmileyCraft had the same idea
$endgroup$
– Omnomnomnom
Dec 20 '18 at 18:49
$begingroup$
Looks like SmileyCraft had the same idea
$endgroup$
– Omnomnomnom
Dec 20 '18 at 18:49
add a comment |
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$begingroup$
What about a continuous surjections that is constant on an interval? for example,$f(x)=x$ for $xle 0$, $f(x)=0$ for $0<x<1$ and $f(x)=x-1$ for $xge 1$.
$endgroup$
– Tito Eliatron
Dec 20 '18 at 18:42
$begingroup$
This is very helpful! Thanks!
$endgroup$
– Lucas Corrêa
Dec 20 '18 at 18:46
2
$begingroup$
If you want one with a closed formula, you can take $f(x)=x^3-x$. Plotting this should make the properties clear.
$endgroup$
– SmileyCraft
Dec 20 '18 at 18:47