Proving a Fixed Point Theorem for Shrinking Maps on Compact Spaces

[facenian]

TL;DR

A problem in Munkres' topology book $\S$-7(b) Page 183

Show that if $f$ is a shrinking map $d(f(x),f(y)) < d(x,y)$ and $X$ is compact, then $f$ has a unique fixed point.
Hint. Let $A_n=f^n(X)$ and $A=\cap A_n$. Given $x\in A$, choose $x_n$ so that $x=f^{n+1}(x_n)$. If $a$ is the limit of some subsequence of the sequence $y_n=f^n(x_n)$, show that $a\in A$ and $f(a)=x$. Conlude that $A=f(A)$, so that $diam\,A=0$.

Solution: I can prove all except that $a\in A$, i.e., I had to assume it to prove that $A=f(A)$ and $diam\,A=0$.
Any help will be welcome.

 

[member 587159]

I will not bother working with a subsequence, and I'll take the full sequence. The easy modification to the general case is left to you.

You should show that $a \in A_m$ for all $m \geq 1$. Note that we have a descending chain $A_1 \supseteq A_2 \supseteq \dots$. Fix $m \geq 1$.

Observe that $y_n \in A_n \subseteq A_m$ for $n \geq m$. Thus

$$a = \lim_{n \to \infty, n \geq m} y_n \in A_m$$ because $A_m$ is closed (limits of sequences in a set remain in the closure of the set).

Since $m \geq 1$ is arbitrary, $a \in \bigcap_{m=1}^\infty A_m=A$.

 

[facenian]

I will not bother working with a subsequence, and I'll take the full sequence. The easy modification to the general case is left to you.

You should show that $a \in A_m$ for all $m \geq 1$. Note that we have a descending chain $A_1 \supseteq A_2 \supseteq \dots$. Fix $m \geq 1$.

Observe that $y_n \in A_n \subseteq A_m$ for $n \geq m$. Thus

$$a = \lim_{n \to \infty, n \geq m} y_n \in A_m$$ because $A_m$ is closed (limits of sequences in a set remain in the closure of the set).

Since $m \geq 1$ is arbitrary, $a \in \bigcap_{m=1}^\infty A_m=A$.

Many thanks for your response, however, I believe that it is not correct. Let us just work with the sequence as you said. Your answer would be correct if $y_n\in A$ then $a$ would be in $\overline{A}=A$. The problem is that $y_n\not\in A$ in general because we cannot assume that $y_n\in A_m$ for $m>n$, so $a$ is a limit point for the elements $\{y_k:k\in Z^+\}$ but not for elements in $A$.

 

[member 587159]

Many thanks for your response, however, I believe that it is not correct. Let us just work with the sequence as you said. Your answer would be correct if $y_n\in A$ then $a$ would be in $\overline{A}=A$. The problem is that $y_n\not\in A$ in general because we cannot assume that $y_n\in A_m$ for $m>n$, so $a$ is a limit point for the elements $\{y_k:k\in Z^+\}$ but not for elements in $A$.

I never claim that $y_n\in A$. Reread my answer more carefully please.

I don't show directly that $a\in A$, I rather show that $a \in A_m$ for every $m$ and this follows because the sequence lives in $A_m$ for sufficiently large indices. Hence the limit lives in $\overline{A_m}=A_m$.

You can do this for every $m$ and then you can conclude that $a \in A$.

Could you let me know if this helped?

 

[facenian]

I never claim that $y_n\in A$. Reread my answer more carefully please.

I don't show directly that $a\in A$, I rather show that $a \in A_m$ for every $m$ and this follows because the sequence lives in $A_m$ for sufficiently large indices. Hence the limit lives in $\overline{A_m}=A_m$.

You can do this for every $m$ and then you can conclude that $a \in A$.

You are right! I read it and interpreted it as I tried to solve it before. Thank you.

 

[WWGD]

Each An is compact and therefore closed ( in a metric space) and thus so is A as the intersection, and complete ( compact subset/subspace of metric is complete).A Cauchy sequence in A must therefore converge to a point in A.

 

[member 587159]

Each An is compact and therefore closed ( in a metric space) and thus so is A as the intersection, and complete ( compact subset/subspace of metric is complete).A Cauchy sequence in A must therefore converge to a point in A.

The sequence does not live in $A$.

 

[WWGD]

Yes, I clearly oversped on this one. Let me read more carefully.

 

[wrobel]

Let $(X,d)$ be a metric space and let $f:X\to X$ be such that
$$d(f(x),f(y))<d(x,y),\quad x\ne y.$$

Theorem. Assume that for some $x_0$ a sequence $x_{n+1}=f(x_n)$ contains a convergent subsequence: $x_{n_k}\to x_*$. Then $f(x_*)=x_*$