# How to show there exist a unique $x_o$ in X such that f($x_o$)=$x_o$? [duplicate]

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.

• 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. – ccroth Jun 24 at 15:29
• Try checking out the Brouwer fixed-point theorem en.wikipedia.org/wiki/Brouwer_fixed-point_theorem – Zim Jun 24 at 15:32
• Sorry sir actually edit option is not showing and i dont know why. But yes i mean that. – Sunit das Jun 24 at 15:32
• @TravisWillse Banach fixed point theorem requires a contraction, and this is not. Note that without compactness, it is easy to construct counterexamples. – N. S. Jun 24 at 15:50
• @N.S. Oops, you're right, I misread the condition in O.P.'s statement of the problem. Thanks for pointing it out. – Travis Willse Jun 24 at 15:56

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 $$d(f(x),f(y)) implies that the image $$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).
• 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? – N. S. Jun 24 at 15:55