Continuous map from a hausdorff space to itself
$begingroup$
Let X be a compact Hausdorff space, and let $f:X to X$ be a continuous map of X into itself. Prove that there is a non-empty subset A of X such that $f(A)=A$.
If I let $A_1 = f(X)$ and inductively $A_{n+1}=f(A_n)$. Let $A=cap _{n=1}^infty A_n $
Why does this work?
analysis
asked Dec 5 '18 at 6:16
general1597general1597
313
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add a comment |
$begingroup$
Let X be a compact Hausdorff space, and let $f:X to X$ be a continuous map of X into itself. Prove that there is a non-empty subset A of X such that $f(A)=A$.
If I let $A_1 = f(X)$ and inductively $A_{n+1}=f(A_n)$. Let $A=cap _{n=1}^infty A_n $
Why does this work?
analysis
asked Dec 5 '18 at 6:16
general1597general1597
313
$endgroup$
add a comment |
$begingroup$
Let X be a compact Hausdorff space, and let $f:X to X$ be a continuous map of X into itself. Prove that there is a non-empty subset A of X such that $f(A)=A$.
If I let $A_1 = f(X)$ and inductively $A_{n+1}=f(A_n)$. Let $A=cap _{n=1}^infty A_n $
Why does this work?
analysis
asked Dec 5 '18 at 6:16
general1597general1597
313
$endgroup$
Let X be a compact Hausdorff space, and let $f:X to X$ be a continuous map of X into itself. Prove that there is a non-empty subset A of X such that $f(A)=A$.
If I let $A_1 = f(X)$ and inductively $A_{n+1}=f(A_n)$. Let $A=cap _{n=1}^infty A_n $
Why does this work?
analysis
analysis
asked Dec 5 '18 at 6:16
general1597general1597
313
asked Dec 5 '18 at 6:16
general1597general1597
313
asked Dec 5 '18 at 6:16
general1597general1597
313
asked Dec 5 '18 at 6:16
general1597general1597
313
asked Dec 5 '18 at 6:16
general1597general1597
313
313
add a comment |
add a comment |
2 Answers
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Prove by induction that $A_{n+1} subset A_n$. Let $y in A$. Then for each $n$, $y in A_{n+1}$ so $y=f(x_n)$ with $x_n in A_n$. By compactness there is a subnet of ${x_n}$ which converges to some point $x$. Note that $x_{n+m} in A_{n+m} subset A_n$ so $x in A_n$ for each $n$. [ Of course the sets $A_n$ are all compact, hence closed]. By continuity of $f$ we also have $y=f(x)$. Hence $y in f(A)$. We have proved that $A subset f(A)$. The reverse inclusion is easy. If more details are needed let me know.
answered Dec 5 '18 at 6:35
Kavi Rama MurthyKavi Rama Murthy
52.9k32055
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add a comment |
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Nitpicking 101: The empty space is compact Hausdorff. So assume $Xne emptyset.$
Let $X_0=X$ and $X_{n+1}=f(X_n)$ for $ngeq 0.$ Obviously $A_1subset A_0.$ If $A_{n+1}subset A_n$ then $$A_{n+2}=f(A_{n+1})=f(A_{n+1}cap A_n)subset f(A_n)=A_{n+1}.$$ So by induction, $A_{n+1}subset A_n$ for all $ngeq 0.$
Any statement about open sets has an equivalent "dual" statement about closed sets. An equivalent to "Every open cover of $X$ has a finite sub-cover" is:
$(bullet).$ If $F$ is a non-empty closed family in $X$ and $F$ has the FIP then $cap Fne emptyset.$
Translation: A closed family in $X$ is a family of closed subsets of $X...$ FIP is Finite Intersection Property, which means that if $G$ is a non-empty finite subset of $F$ then $cap Gne emptyset.$
Let $F={A_n:ngeq 0}$. Assuming $Xne emptyset,$ we have (fairly obviously) $A_nne emptysetimplies A_{n+1}= f(A_n)ne emptyset,$ so $F$ has the FIP. Now the continuous image of a compact space is compact, so by induction on $ngeq 0,$ every $A_n$ is compact. And since $X$ is Hausdorff, compact sub-spaces of $X$ are closed in $X$. Hence by $(bullet )$ we have $A=cap Fne emptyset.$
It should be fairly obvious that $f(A)subset A.$ To show that $Asubset f(A),$ take any $ain A,$ and for $ngeq 0$ let $H_{n,a}=A_ncap f^{-1}{a}.$ We have $$ain A_{n+1}=f(A_n)implies emptyset ne A_n cap f^{-1}{a}=H_{n,a}.$$ Now $H_a={H_{n,a}: ngeq 0}$ is a non-empty closed family in $X$ with the FIP so by $(bullet )$ we have $cap H_a ne emptyset.$ So take any $bin cap H_a.$ Then $f(b)=a$ and $bin cap_{ngeq 0}A_n=A,$ so $ain f(A).$
Remark: ${a}$ is closed because $X$ is Hausdorff. And $f$ is continuous so $f^{-1}C$ is closed when $C$ is closed . So $f^{-1}{a}$ is closed. And every $A_n$ is closed. So $H_{n,a}$ is closed.
answered Dec 5 '18 at 9:31
DanielWainfleetDanielWainfleet
34.6k31648
$endgroup$
$begingroup$
I found this to longer and subtler than I expected, especially that $Asubset f(A)$.
$endgroup$
– DanielWainfleet
Dec 5 '18 at 9:33
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Prove by induction that $A_{n+1} subset A_n$. Let $y in A$. Then for each $n$, $y in A_{n+1}$ so $y=f(x_n)$ with $x_n in A_n$. By compactness there is a subnet of ${x_n}$ which converges to some point $x$. Note that $x_{n+m} in A_{n+m} subset A_n$ so $x in A_n$ for each $n$. [ Of course the sets $A_n$ are all compact, hence closed]. By continuity of $f$ we also have $y=f(x)$. Hence $y in f(A)$. We have proved that $A subset f(A)$. The reverse inclusion is easy. If more details are needed let me know.
answered Dec 5 '18 at 6:35
Kavi Rama MurthyKavi Rama Murthy
52.9k32055
$endgroup$
add a comment |
$begingroup$
Prove by induction that $A_{n+1} subset A_n$. Let $y in A$. Then for each $n$, $y in A_{n+1}$ so $y=f(x_n)$ with $x_n in A_n$. By compactness there is a subnet of ${x_n}$ which converges to some point $x$. Note that $x_{n+m} in A_{n+m} subset A_n$ so $x in A_n$ for each $n$. [ Of course the sets $A_n$ are all compact, hence closed]. By continuity of $f$ we also have $y=f(x)$. Hence $y in f(A)$. We have proved that $A subset f(A)$. The reverse inclusion is easy. If more details are needed let me know.
answered Dec 5 '18 at 6:35
Kavi Rama MurthyKavi Rama Murthy
52.9k32055
$endgroup$
add a comment |
$begingroup$
Prove by induction that $A_{n+1} subset A_n$. Let $y in A$. Then for each $n$, $y in A_{n+1}$ so $y=f(x_n)$ with $x_n in A_n$. By compactness there is a subnet of ${x_n}$ which converges to some point $x$. Note that $x_{n+m} in A_{n+m} subset A_n$ so $x in A_n$ for each $n$. [ Of course the sets $A_n$ are all compact, hence closed]. By continuity of $f$ we also have $y=f(x)$. Hence $y in f(A)$. We have proved that $A subset f(A)$. The reverse inclusion is easy. If more details are needed let me know.
answered Dec 5 '18 at 6:35
Kavi Rama MurthyKavi Rama Murthy
52.9k32055
$endgroup$
Prove by induction that $A_{n+1} subset A_n$. Let $y in A$. Then for each $n$, $y in A_{n+1}$ so $y=f(x_n)$ with $x_n in A_n$. By compactness there is a subnet of ${x_n}$ which converges to some point $x$. Note that $x_{n+m} in A_{n+m} subset A_n$ so $x in A_n$ for each $n$. [ Of course the sets $A_n$ are all compact, hence closed]. By continuity of $f$ we also have $y=f(x)$. Hence $y in f(A)$. We have proved that $A subset f(A)$. The reverse inclusion is easy. If more details are needed let me know.
answered Dec 5 '18 at 6:35
Kavi Rama MurthyKavi Rama Murthy
52.9k32055
answered Dec 5 '18 at 6:35
Kavi Rama MurthyKavi Rama Murthy
52.9k32055
answered Dec 5 '18 at 6:35
Kavi Rama MurthyKavi Rama Murthy
52.9k32055
answered Dec 5 '18 at 6:35
Kavi Rama MurthyKavi Rama Murthy
52.9k32055
52.9k32055
add a comment |
add a comment |
$begingroup$
Nitpicking 101: The empty space is compact Hausdorff. So assume $Xne emptyset.$
Let $X_0=X$ and $X_{n+1}=f(X_n)$ for $ngeq 0.$ Obviously $A_1subset A_0.$ If $A_{n+1}subset A_n$ then $$A_{n+2}=f(A_{n+1})=f(A_{n+1}cap A_n)subset f(A_n)=A_{n+1}.$$ So by induction, $A_{n+1}subset A_n$ for all $ngeq 0.$
Any statement about open sets has an equivalent "dual" statement about closed sets. An equivalent to "Every open cover of $X$ has a finite sub-cover" is:
$(bullet).$ If $F$ is a non-empty closed family in $X$ and $F$ has the FIP then $cap Fne emptyset.$
Translation: A closed family in $X$ is a family of closed subsets of $X...$ FIP is Finite Intersection Property, which means that if $G$ is a non-empty finite subset of $F$ then $cap Gne emptyset.$
Let $F={A_n:ngeq 0}$. Assuming $Xne emptyset,$ we have (fairly obviously) $A_nne emptysetimplies A_{n+1}= f(A_n)ne emptyset,$ so $F$ has the FIP. Now the continuous image of a compact space is compact, so by induction on $ngeq 0,$ every $A_n$ is compact. And since $X$ is Hausdorff, compact sub-spaces of $X$ are closed in $X$. Hence by $(bullet )$ we have $A=cap Fne emptyset.$
It should be fairly obvious that $f(A)subset A.$ To show that $Asubset f(A),$ take any $ain A,$ and for $ngeq 0$ let $H_{n,a}=A_ncap f^{-1}{a}.$ We have $$ain A_{n+1}=f(A_n)implies emptyset ne A_n cap f^{-1}{a}=H_{n,a}.$$ Now $H_a={H_{n,a}: ngeq 0}$ is a non-empty closed family in $X$ with the FIP so by $(bullet )$ we have $cap H_a ne emptyset.$ So take any $bin cap H_a.$ Then $f(b)=a$ and $bin cap_{ngeq 0}A_n=A,$ so $ain f(A).$
Remark: ${a}$ is closed because $X$ is Hausdorff. And $f$ is continuous so $f^{-1}C$ is closed when $C$ is closed . So $f^{-1}{a}$ is closed. And every $A_n$ is closed. So $H_{n,a}$ is closed.
answered Dec 5 '18 at 9:31
DanielWainfleetDanielWainfleet
34.6k31648
$endgroup$
$begingroup$
I found this to longer and subtler than I expected, especially that $Asubset f(A)$.
$endgroup$
– DanielWainfleet
Dec 5 '18 at 9:33
add a comment |
$begingroup$
Nitpicking 101: The empty space is compact Hausdorff. So assume $Xne emptyset.$
Let $X_0=X$ and $X_{n+1}=f(X_n)$ for $ngeq 0.$ Obviously $A_1subset A_0.$ If $A_{n+1}subset A_n$ then $$A_{n+2}=f(A_{n+1})=f(A_{n+1}cap A_n)subset f(A_n)=A_{n+1}.$$ So by induction, $A_{n+1}subset A_n$ for all $ngeq 0.$
Any statement about open sets has an equivalent "dual" statement about closed sets. An equivalent to "Every open cover of $X$ has a finite sub-cover" is:
$(bullet).$ If $F$ is a non-empty closed family in $X$ and $F$ has the FIP then $cap Fne emptyset.$
Translation: A closed family in $X$ is a family of closed subsets of $X...$ FIP is Finite Intersection Property, which means that if $G$ is a non-empty finite subset of $F$ then $cap Gne emptyset.$
Let $F={A_n:ngeq 0}$. Assuming $Xne emptyset,$ we have (fairly obviously) $A_nne emptysetimplies A_{n+1}= f(A_n)ne emptyset,$ so $F$ has the FIP. Now the continuous image of a compact space is compact, so by induction on $ngeq 0,$ every $A_n$ is compact. And since $X$ is Hausdorff, compact sub-spaces of $X$ are closed in $X$. Hence by $(bullet )$ we have $A=cap Fne emptyset.$
It should be fairly obvious that $f(A)subset A.$ To show that $Asubset f(A),$ take any $ain A,$ and for $ngeq 0$ let $H_{n,a}=A_ncap f^{-1}{a}.$ We have $$ain A_{n+1}=f(A_n)implies emptyset ne A_n cap f^{-1}{a}=H_{n,a}.$$ Now $H_a={H_{n,a}: ngeq 0}$ is a non-empty closed family in $X$ with the FIP so by $(bullet )$ we have $cap H_a ne emptyset.$ So take any $bin cap H_a.$ Then $f(b)=a$ and $bin cap_{ngeq 0}A_n=A,$ so $ain f(A).$
Remark: ${a}$ is closed because $X$ is Hausdorff. And $f$ is continuous so $f^{-1}C$ is closed when $C$ is closed . So $f^{-1}{a}$ is closed. And every $A_n$ is closed. So $H_{n,a}$ is closed.
answered Dec 5 '18 at 9:31
DanielWainfleetDanielWainfleet
34.6k31648
$endgroup$
$begingroup$
I found this to longer and subtler than I expected, especially that $Asubset f(A)$.
$endgroup$
– DanielWainfleet
Dec 5 '18 at 9:33
add a comment |
$begingroup$
Nitpicking 101: The empty space is compact Hausdorff. So assume $Xne emptyset.$
Let $X_0=X$ and $X_{n+1}=f(X_n)$ for $ngeq 0.$ Obviously $A_1subset A_0.$ If $A_{n+1}subset A_n$ then $$A_{n+2}=f(A_{n+1})=f(A_{n+1}cap A_n)subset f(A_n)=A_{n+1}.$$ So by induction, $A_{n+1}subset A_n$ for all $ngeq 0.$
Any statement about open sets has an equivalent "dual" statement about closed sets. An equivalent to "Every open cover of $X$ has a finite sub-cover" is:
$(bullet).$ If $F$ is a non-empty closed family in $X$ and $F$ has the FIP then $cap Fne emptyset.$
Translation: A closed family in $X$ is a family of closed subsets of $X...$ FIP is Finite Intersection Property, which means that if $G$ is a non-empty finite subset of $F$ then $cap Gne emptyset.$
Let $F={A_n:ngeq 0}$. Assuming $Xne emptyset,$ we have (fairly obviously) $A_nne emptysetimplies A_{n+1}= f(A_n)ne emptyset,$ so $F$ has the FIP. Now the continuous image of a compact space is compact, so by induction on $ngeq 0,$ every $A_n$ is compact. And since $X$ is Hausdorff, compact sub-spaces of $X$ are closed in $X$. Hence by $(bullet )$ we have $A=cap Fne emptyset.$
It should be fairly obvious that $f(A)subset A.$ To show that $Asubset f(A),$ take any $ain A,$ and for $ngeq 0$ let $H_{n,a}=A_ncap f^{-1}{a}.$ We have $$ain A_{n+1}=f(A_n)implies emptyset ne A_n cap f^{-1}{a}=H_{n,a}.$$ Now $H_a={H_{n,a}: ngeq 0}$ is a non-empty closed family in $X$ with the FIP so by $(bullet )$ we have $cap H_a ne emptyset.$ So take any $bin cap H_a.$ Then $f(b)=a$ and $bin cap_{ngeq 0}A_n=A,$ so $ain f(A).$
Remark: ${a}$ is closed because $X$ is Hausdorff. And $f$ is continuous so $f^{-1}C$ is closed when $C$ is closed . So $f^{-1}{a}$ is closed. And every $A_n$ is closed. So $H_{n,a}$ is closed.
answered Dec 5 '18 at 9:31
DanielWainfleetDanielWainfleet
34.6k31648
$endgroup$
Nitpicking 101: The empty space is compact Hausdorff. So assume $Xne emptyset.$
Let $X_0=X$ and $X_{n+1}=f(X_n)$ for $ngeq 0.$ Obviously $A_1subset A_0.$ If $A_{n+1}subset A_n$ then $$A_{n+2}=f(A_{n+1})=f(A_{n+1}cap A_n)subset f(A_n)=A_{n+1}.$$ So by induction, $A_{n+1}subset A_n$ for all $ngeq 0.$
Any statement about open sets has an equivalent "dual" statement about closed sets. An equivalent to "Every open cover of $X$ has a finite sub-cover" is:
$(bullet).$ If $F$ is a non-empty closed family in $X$ and $F$ has the FIP then $cap Fne emptyset.$
Translation: A closed family in $X$ is a family of closed subsets of $X...$ FIP is Finite Intersection Property, which means that if $G$ is a non-empty finite subset of $F$ then $cap Gne emptyset.$
Let $F={A_n:ngeq 0}$. Assuming $Xne emptyset,$ we have (fairly obviously) $A_nne emptysetimplies A_{n+1}= f(A_n)ne emptyset,$ so $F$ has the FIP. Now the continuous image of a compact space is compact, so by induction on $ngeq 0,$ every $A_n$ is compact. And since $X$ is Hausdorff, compact sub-spaces of $X$ are closed in $X$. Hence by $(bullet )$ we have $A=cap Fne emptyset.$
It should be fairly obvious that $f(A)subset A.$ To show that $Asubset f(A),$ take any $ain A,$ and for $ngeq 0$ let $H_{n,a}=A_ncap f^{-1}{a}.$ We have $$ain A_{n+1}=f(A_n)implies emptyset ne A_n cap f^{-1}{a}=H_{n,a}.$$ Now $H_a={H_{n,a}: ngeq 0}$ is a non-empty closed family in $X$ with the FIP so by $(bullet )$ we have $cap H_a ne emptyset.$ So take any $bin cap H_a.$ Then $f(b)=a$ and $bin cap_{ngeq 0}A_n=A,$ so $ain f(A).$
Remark: ${a}$ is closed because $X$ is Hausdorff. And $f$ is continuous so $f^{-1}C$ is closed when $C$ is closed . So $f^{-1}{a}$ is closed. And every $A_n$ is closed. So $H_{n,a}$ is closed.
answered Dec 5 '18 at 9:31
DanielWainfleetDanielWainfleet
34.6k31648
answered Dec 5 '18 at 9:31
DanielWainfleetDanielWainfleet
34.6k31648
answered Dec 5 '18 at 9:31
DanielWainfleetDanielWainfleet
34.6k31648
answered Dec 5 '18 at 9:31
DanielWainfleetDanielWainfleet
34.6k31648
34.6k31648
$begingroup$
I found this to longer and subtler than I expected, especially that $Asubset f(A)$.
$endgroup$
– DanielWainfleet
Dec 5 '18 at 9:33
add a comment |
$begingroup$
I found this to longer and subtler than I expected, especially that $Asubset f(A)$.
$endgroup$
– DanielWainfleet
Dec 5 '18 at 9:33
$begingroup$
I found this to longer and subtler than I expected, especially that $Asubset f(A)$.
$endgroup$
– DanielWainfleet
Dec 5 '18 at 9:33
$begingroup$
I found this to longer and subtler than I expected, especially that $Asubset f(A)$.
$endgroup$
– DanielWainfleet
Dec 5 '18 at 9:33
add a comment |
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