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Let $(X, d)$ be a compact metric space. Let $f:X\rightarrow X$ be such that $d(f(x), f(y)) < d(x, y)$ for all $x, y\in X$ with $x$ not equal to $y$. Show that $f$ has a fixed point, that is, there exists $x_0\in X$ such that $f(x_0) = x_0$. Is the fixed point unique?

My work: first i prove that $f$ is uniformly continuous on $X$ and if possible $f(x) \neq x$ for all $x \in X$ Considering a function $x\rightarrow d(x,f(x))$. Then i showed that this function is continuous by sequential criteria of continuity. Since it is continuous on a metric space then it is also uniformly continuous and attains it's infimum at some point $x_1 \in X$ so $d(x_1,f(x_1))>0$. But after that I cannot proceed. I think it will contradict the given condition. But ran out of ideas how to show that.

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  • $\begingroup$ Submitted an edit to try cleaning up the formatting. Your work is a bit unclear, I think you mean to say "suppose $f(x) \neq x$ for all $x \in X$" in the beginning. $\endgroup$ – ccroth Jun 24 at 15:29
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    $\begingroup$ Try checking out the Brouwer fixed-point theorem en.wikipedia.org/wiki/Brouwer_fixed-point_theorem $\endgroup$ – Zim Jun 24 at 15:32
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    $\begingroup$ Sorry sir actually edit option is not showing and i dont know why. But yes i mean that. $\endgroup$ – Sunit das Jun 24 at 15:32
  • $\begingroup$ @TravisWillse Banach fixed point theorem requires a contraction, and this is not. Note that without compactness, it is easy to construct counterexamples. $\endgroup$ – N. S. Jun 24 at 15:50
  • $\begingroup$ @N.S. Oops, you're right, I misread the condition in O.P.'s statement of the problem. Thanks for pointing it out. $\endgroup$ – Travis Willse Jun 24 at 15:56
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We can actually construct a fixed point by repeatedly applying $f$ to the set $X$ and taking the limit. Here‘s a sketch.

Let $Y\subset X$. The condition that $d(f(x),f(y))<d(x,y)$ implies that the image $f(Y)$ of $Y$ cannot equal $Y$ (the diameter is finite because the set is bounded, and the diameter must shrink when $f$ is applied). Therefore we have $f(Y)\subset Y$ for any subset $Y$ of $X$.

This implies that $$X\supset f(X)\supset f(f(X))\supset f(f(f(X)))\supset ...$$ In this sequence of sets, let us denote the nth set by $X_n$ so that $X_0=X$, $X_1=f(X)$, and so on. We have that $X_{n+1}\subset X_n$, and all $X_n$ are nonempty.

Notice that any point inside of the set $$\bigcap_{n=0}^\infty X_n$$ is a fixed-point of the function $f$. We should be able to construct such a point by taking the limit of a converging sequence of points $(x_0,x_1,x_2,...)$ with $x_n\in X_n$. We know that the limit point will be an element of $X$ because it is compact (and therefore closed).

Did I miss anything? I may have skipped some steps, so maybe people can help me clean up this proof in the comments.

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  • $\begingroup$ I think it is not true that $f(Y) \subset Y$ for all $Y \subset X$ (just think about $Y$ being a single point). On another hand it is easy to see that $X_{n+1} \subseteq X_n$. Also, the line "notice that any point inside the set is a fixed point" is not clear, why is this the case? $\endgroup$ – N. S. Jun 24 at 15:55

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