Does there exist a function such that the preimage of $x ^ { 2 } + y ^ { 2 } leq 1$ is the closed interval...












1












$begingroup$


Does there exist a continuous function $f : mathbb { R } rightarrow mathbb { R } ^ { 2 }$ such
that the preimage of the closed unit disk $x ^ { 2 } + y ^ { 2 } leq 1$ is the closed
interval $[ - 1,1 ] ?$ the open interval $( - 1,1 ) ?$



To be honest, I don't really know how to go about this problem.






real-analysis functions continuity







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$endgroup$












  • $begingroup$
    en.wikipedia.org/wiki/Space-filling_curve
    $endgroup$
    – Will Jagy
    Dec 20 '18 at 1:39










  • $begingroup$
    I'm fairly certain you can get a continuous function from $[-1,1]$ to the closed unit disc by an appropriate space filling curve.
    $endgroup$
    – MathematicsStudent1122
    Dec 20 '18 at 1:39








  • 2




    $begingroup$
    the preimage of a set $S$ being $[-1,1]$ is not the same thing as the image of $[-1,1]$ being $S$. space filling curves are super overkill
    $endgroup$
    – Rolf Hoyer
    Dec 20 '18 at 1:41












  • $begingroup$
    @RolfHoyer how would you go about this without using space filling functions? We never went over those in my lecture so I doubt the answer would have to be related to that.
    $endgroup$
    – Mohammed Shahid
    Dec 20 '18 at 2:14
















1












$begingroup$


Does there exist a continuous function $f : mathbb { R } rightarrow mathbb { R } ^ { 2 }$ such
that the preimage of the closed unit disk $x ^ { 2 } + y ^ { 2 } leq 1$ is the closed
interval $[ - 1,1 ] ?$ the open interval $( - 1,1 ) ?$



To be honest, I don't really know how to go about this problem.






real-analysis functions continuity







share|cite|improve this question









$endgroup$












  • $begingroup$
    en.wikipedia.org/wiki/Space-filling_curve
    $endgroup$
    – Will Jagy
    Dec 20 '18 at 1:39










  • $begingroup$
    I'm fairly certain you can get a continuous function from $[-1,1]$ to the closed unit disc by an appropriate space filling curve.
    $endgroup$
    – MathematicsStudent1122
    Dec 20 '18 at 1:39








  • 2




    $begingroup$
    the preimage of a set $S$ being $[-1,1]$ is not the same thing as the image of $[-1,1]$ being $S$. space filling curves are super overkill
    $endgroup$
    – Rolf Hoyer
    Dec 20 '18 at 1:41












  • $begingroup$
    @RolfHoyer how would you go about this without using space filling functions? We never went over those in my lecture so I doubt the answer would have to be related to that.
    $endgroup$
    – Mohammed Shahid
    Dec 20 '18 at 2:14














1












1








1


1



$begingroup$


Does there exist a continuous function $f : mathbb { R } rightarrow mathbb { R } ^ { 2 }$ such
that the preimage of the closed unit disk $x ^ { 2 } + y ^ { 2 } leq 1$ is the closed
interval $[ - 1,1 ] ?$ the open interval $( - 1,1 ) ?$



To be honest, I don't really know how to go about this problem.






real-analysis functions continuity







share|cite|improve this question









$endgroup$




Does there exist a continuous function $f : mathbb { R } rightarrow mathbb { R } ^ { 2 }$ such
that the preimage of the closed unit disk $x ^ { 2 } + y ^ { 2 } leq 1$ is the closed
interval $[ - 1,1 ] ?$ the open interval $( - 1,1 ) ?$



To be honest, I don't really know how to go about this problem.






real-analysis functions continuity




real-analysis functions continuity






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 20 '18 at 1:29









Mohammed ShahidMohammed Shahid

1457




1457












  • $begingroup$
    en.wikipedia.org/wiki/Space-filling_curve
    $endgroup$
    – Will Jagy
    Dec 20 '18 at 1:39










  • $begingroup$
    I'm fairly certain you can get a continuous function from $[-1,1]$ to the closed unit disc by an appropriate space filling curve.
    $endgroup$
    – MathematicsStudent1122
    Dec 20 '18 at 1:39








  • 2




    $begingroup$
    the preimage of a set $S$ being $[-1,1]$ is not the same thing as the image of $[-1,1]$ being $S$. space filling curves are super overkill
    $endgroup$
    – Rolf Hoyer
    Dec 20 '18 at 1:41












  • $begingroup$
    @RolfHoyer how would you go about this without using space filling functions? We never went over those in my lecture so I doubt the answer would have to be related to that.
    $endgroup$
    – Mohammed Shahid
    Dec 20 '18 at 2:14


















  • $begingroup$
    en.wikipedia.org/wiki/Space-filling_curve
    $endgroup$
    – Will Jagy
    Dec 20 '18 at 1:39










  • $begingroup$
    I'm fairly certain you can get a continuous function from $[-1,1]$ to the closed unit disc by an appropriate space filling curve.
    $endgroup$
    – MathematicsStudent1122
    Dec 20 '18 at 1:39








  • 2




    $begingroup$
    the preimage of a set $S$ being $[-1,1]$ is not the same thing as the image of $[-1,1]$ being $S$. space filling curves are super overkill
    $endgroup$
    – Rolf Hoyer
    Dec 20 '18 at 1:41












  • $begingroup$
    @RolfHoyer how would you go about this without using space filling functions? We never went over those in my lecture so I doubt the answer would have to be related to that.
    $endgroup$
    – Mohammed Shahid
    Dec 20 '18 at 2:14
















$begingroup$
en.wikipedia.org/wiki/Space-filling_curve
$endgroup$
– Will Jagy
Dec 20 '18 at 1:39




$begingroup$
en.wikipedia.org/wiki/Space-filling_curve
$endgroup$
– Will Jagy
Dec 20 '18 at 1:39












$begingroup$
I'm fairly certain you can get a continuous function from $[-1,1]$ to the closed unit disc by an appropriate space filling curve.
$endgroup$
– MathematicsStudent1122
Dec 20 '18 at 1:39






$begingroup$
I'm fairly certain you can get a continuous function from $[-1,1]$ to the closed unit disc by an appropriate space filling curve.
$endgroup$
– MathematicsStudent1122
Dec 20 '18 at 1:39






2




2




$begingroup$
the preimage of a set $S$ being $[-1,1]$ is not the same thing as the image of $[-1,1]$ being $S$. space filling curves are super overkill
$endgroup$
– Rolf Hoyer
Dec 20 '18 at 1:41






$begingroup$
the preimage of a set $S$ being $[-1,1]$ is not the same thing as the image of $[-1,1]$ being $S$. space filling curves are super overkill
$endgroup$
– Rolf Hoyer
Dec 20 '18 at 1:41














$begingroup$
@RolfHoyer how would you go about this without using space filling functions? We never went over those in my lecture so I doubt the answer would have to be related to that.
$endgroup$
– Mohammed Shahid
Dec 20 '18 at 2:14




$begingroup$
@RolfHoyer how would you go about this without using space filling functions? We never went over those in my lecture so I doubt the answer would have to be related to that.
$endgroup$
– Mohammed Shahid
Dec 20 '18 at 2:14










2 Answers
2






active

oldest

votes


















3












$begingroup$

Space filling curves certainly work but that's quite excessive.



For a function $f:Bbb RtoBbb R^2$ defined by $f(x):=(f_1(x),f_2(x))$, the preimage of $D:={(x,y)inBbb R^2:x^2+y^2le 1 }$ is merely the set
$$begin{align}
f^{-1}(D) &= { xinBbb R: (f_1(x),f_2(x))in D } \
&= { xinBbb R: f_1(x)^2+f_2(x)^2le 1 }.
end{align}$$



By letting $f(x)=(x,0)$, we have
$$
f^{-1}(D)= { xinBbb R: x^2le 1 } = [-1,1].
$$



On the other hand, by requiring that $f$ be continuous, the preimage of $D$, which is a closed set, must also be closed. Since $(-1,1)$ is not closed, we cannot find a continuous function such that $f^{-1}(D)=(-1,1)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yep. The trick here is to realize that the preimage need not necessarily give the given set when the function is then applied to it - only that the result of such application be contained within the given set.
    $endgroup$
    – The_Sympathizer
    Dec 20 '18 at 6:13



















1












$begingroup$

f:R -> R×R, x -> (x,0) is continuous.

Let D be the closed unit disk.

The preimage of D by f is

f$^{-1}$(D) = { x : f(x) in D } = [-1,1].






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Good example, but you really should format it better.
    $endgroup$
    – zhw.
    Dec 20 '18 at 17:25











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Space filling curves certainly work but that's quite excessive.



For a function $f:Bbb RtoBbb R^2$ defined by $f(x):=(f_1(x),f_2(x))$, the preimage of $D:={(x,y)inBbb R^2:x^2+y^2le 1 }$ is merely the set
$$begin{align}
f^{-1}(D) &= { xinBbb R: (f_1(x),f_2(x))in D } \
&= { xinBbb R: f_1(x)^2+f_2(x)^2le 1 }.
end{align}$$



By letting $f(x)=(x,0)$, we have
$$
f^{-1}(D)= { xinBbb R: x^2le 1 } = [-1,1].
$$



On the other hand, by requiring that $f$ be continuous, the preimage of $D$, which is a closed set, must also be closed. Since $(-1,1)$ is not closed, we cannot find a continuous function such that $f^{-1}(D)=(-1,1)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yep. The trick here is to realize that the preimage need not necessarily give the given set when the function is then applied to it - only that the result of such application be contained within the given set.
    $endgroup$
    – The_Sympathizer
    Dec 20 '18 at 6:13
















3












$begingroup$

Space filling curves certainly work but that's quite excessive.



For a function $f:Bbb RtoBbb R^2$ defined by $f(x):=(f_1(x),f_2(x))$, the preimage of $D:={(x,y)inBbb R^2:x^2+y^2le 1 }$ is merely the set
$$begin{align}
f^{-1}(D) &= { xinBbb R: (f_1(x),f_2(x))in D } \
&= { xinBbb R: f_1(x)^2+f_2(x)^2le 1 }.
end{align}$$



By letting $f(x)=(x,0)$, we have
$$
f^{-1}(D)= { xinBbb R: x^2le 1 } = [-1,1].
$$



On the other hand, by requiring that $f$ be continuous, the preimage of $D$, which is a closed set, must also be closed. Since $(-1,1)$ is not closed, we cannot find a continuous function such that $f^{-1}(D)=(-1,1)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yep. The trick here is to realize that the preimage need not necessarily give the given set when the function is then applied to it - only that the result of such application be contained within the given set.
    $endgroup$
    – The_Sympathizer
    Dec 20 '18 at 6:13














3












3








3





$begingroup$

Space filling curves certainly work but that's quite excessive.



For a function $f:Bbb RtoBbb R^2$ defined by $f(x):=(f_1(x),f_2(x))$, the preimage of $D:={(x,y)inBbb R^2:x^2+y^2le 1 }$ is merely the set
$$begin{align}
f^{-1}(D) &= { xinBbb R: (f_1(x),f_2(x))in D } \
&= { xinBbb R: f_1(x)^2+f_2(x)^2le 1 }.
end{align}$$



By letting $f(x)=(x,0)$, we have
$$
f^{-1}(D)= { xinBbb R: x^2le 1 } = [-1,1].
$$



On the other hand, by requiring that $f$ be continuous, the preimage of $D$, which is a closed set, must also be closed. Since $(-1,1)$ is not closed, we cannot find a continuous function such that $f^{-1}(D)=(-1,1)$.






share|cite|improve this answer











$endgroup$



Space filling curves certainly work but that's quite excessive.



For a function $f:Bbb RtoBbb R^2$ defined by $f(x):=(f_1(x),f_2(x))$, the preimage of $D:={(x,y)inBbb R^2:x^2+y^2le 1 }$ is merely the set
$$begin{align}
f^{-1}(D) &= { xinBbb R: (f_1(x),f_2(x))in D } \
&= { xinBbb R: f_1(x)^2+f_2(x)^2le 1 }.
end{align}$$



By letting $f(x)=(x,0)$, we have
$$
f^{-1}(D)= { xinBbb R: x^2le 1 } = [-1,1].
$$



On the other hand, by requiring that $f$ be continuous, the preimage of $D$, which is a closed set, must also be closed. Since $(-1,1)$ is not closed, we cannot find a continuous function such that $f^{-1}(D)=(-1,1)$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 20 '18 at 16:55

























answered Dec 20 '18 at 4:43









BigbearZzzBigbearZzz

8,93521652




8,93521652












  • $begingroup$
    Yep. The trick here is to realize that the preimage need not necessarily give the given set when the function is then applied to it - only that the result of such application be contained within the given set.
    $endgroup$
    – The_Sympathizer
    Dec 20 '18 at 6:13


















  • $begingroup$
    Yep. The trick here is to realize that the preimage need not necessarily give the given set when the function is then applied to it - only that the result of such application be contained within the given set.
    $endgroup$
    – The_Sympathizer
    Dec 20 '18 at 6:13
















$begingroup$
Yep. The trick here is to realize that the preimage need not necessarily give the given set when the function is then applied to it - only that the result of such application be contained within the given set.
$endgroup$
– The_Sympathizer
Dec 20 '18 at 6:13




$begingroup$
Yep. The trick here is to realize that the preimage need not necessarily give the given set when the function is then applied to it - only that the result of such application be contained within the given set.
$endgroup$
– The_Sympathizer
Dec 20 '18 at 6:13











1












$begingroup$

f:R -> R×R, x -> (x,0) is continuous.

Let D be the closed unit disk.

The preimage of D by f is

f$^{-1}$(D) = { x : f(x) in D } = [-1,1].






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Good example, but you really should format it better.
    $endgroup$
    – zhw.
    Dec 20 '18 at 17:25
















1












$begingroup$

f:R -> R×R, x -> (x,0) is continuous.

Let D be the closed unit disk.

The preimage of D by f is

f$^{-1}$(D) = { x : f(x) in D } = [-1,1].






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Good example, but you really should format it better.
    $endgroup$
    – zhw.
    Dec 20 '18 at 17:25














1












1








1





$begingroup$

f:R -> R×R, x -> (x,0) is continuous.

Let D be the closed unit disk.

The preimage of D by f is

f$^{-1}$(D) = { x : f(x) in D } = [-1,1].






share|cite|improve this answer









$endgroup$



f:R -> R×R, x -> (x,0) is continuous.

Let D be the closed unit disk.

The preimage of D by f is

f$^{-1}$(D) = { x : f(x) in D } = [-1,1].







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 20 '18 at 2:30









William ElliotWilliam Elliot

8,7962820




8,7962820












  • $begingroup$
    Good example, but you really should format it better.
    $endgroup$
    – zhw.
    Dec 20 '18 at 17:25


















  • $begingroup$
    Good example, but you really should format it better.
    $endgroup$
    – zhw.
    Dec 20 '18 at 17:25
















$begingroup$
Good example, but you really should format it better.
$endgroup$
– zhw.
Dec 20 '18 at 17:25




$begingroup$
Good example, but you really should format it better.
$endgroup$
– zhw.
Dec 20 '18 at 17:25


















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