For a Compact Metric Space $T: X\to X$ has unique fixed point

Question

given: $(X,d)$ is a compact metric space $T:X\to X$ is such that $d(T(x),T(y))<d(x,y)\ \forall x,y\in X$ with $x\neq y$.

Prove that T has a unique fixed point.

Attempt: I think I can prove Uniqueness:

Consider $T(x)=x$ and $T(y)=y,\ x\neq y$

Then, $d(x,y)=d(T(x),T(y))<d(x,y)$ (contradicting original condition).

However, I am having trouble executing the proof that the fixed point exists.

Answer 1

Hint: let $X_0 = X$ and $X_{i+1} = T(X_i)$. What can you say about $\bigcap_i X_i$?

For any subset $Y$ of $X$, define the diameter $D(Y)$ of $Y$ to be $\sup\{d(x, y) \mathrel{|} x, y \in Y\}$. Then if $Y$ has more than one point, $D(Y) > 0$ and, if $Y$ is also closed (and hence compact), there are $x, y \in Y$ such that $d(x, y) = D(Y)$ (because $d : Y \times Y \to \mathbb{R}$ is a continuous function on a compact space and hence achieves its maximum value). Taking $Y = T(Z)$ for $Z$ any closed subset of $X$, this gives us points $x, y \in Z$ such that $D(T(Z)) = d(T(x), T(y)) < d(x, y) \le D(Z)$. Now with $F = \bigcap_i X_i$, we have that $F$ is non-empty and closed (as the intersection of a collection of non-empty closed subsets of the compact set $X$) and satisfies $T(F) = F$. By the above remarks $D(T(F)) = D(F) < D(F)$, giving a contradiction, unless $F$ contains exactly one point, which is the desired fixed point of $T$.

Answer 2

This may be long but I'm including every conceivable detail.(1) To avoid the trivial case, assume $X \ne \phi$...............(2) By the standard $ \epsilon , \delta$ method we see that $T$ is continuous.................. (3) For any non-empty closed $S \subset X $, let $D(S)= \sup \{d(x,y) : x,y \in S\}$. We have $ D(S)\leq D(X)<\infty$ because any metric on a compact space is a bounded metric......................(4) For non-empty closed $S\subset X$ , if $S$ has at least $2$ members then $D(T(S))<D(S)$. PROOF : If not, let $(x_n)_n$ and $(y_n)_n$ be sequences in $S$ with $\lim_{n \to \infty} \ d(T(x_n),T(y_n))= D(S)$.By selecting convergent subsequences of $(x_n)_n$ and $(y_n)_n$ we may suppose WLOG (without loss of generality) that $(x_n)_n$ converges to $x$ and $(y_n)_n$ converges to $y$ with $x, y \in S$ (because $S$ is closed). But by continuity of $T$, we have $0 \ne D(S)= \lim_{n \to \infty}d(T(x_n),T(y_n))=d(T(x),T(y))<d(x,y) \le D(S)$, a contradiction..........................(5)As Rob Arthan suggests, let $X_0=X$ and $X_{n+1}=T(X_n)$. The continuous image of a compact space is compact, so if $X_n$ is closed, it is compact, so $X_{n+1}=T(X_n)$ is compact, hence $X_{n+1}$ is closed.By induction every $T_n$ is closed.We also have $X_n \subset X_{n-1}$ for every positive integer $n$ by induction, for if $ X_n \subset X_{n-1}$ then $X_{n+1}=T(X_n)\subset T(X_{n-1})=X_n$ .Obviously $\phi \ne X_n$ for all $n$..........................(6) Since $X$ is compact, the set $ Y=\cap_n X_n \ne \phi$ is non-empty, and also closed................ (7) We have T(Y)=Y. PROOF : Let $y \in Y$ .For each $n$ we have $y \in X_{n+1}=T(X_n)$ so choose $x_n \in X_n$ with $T(x_n)=y$. Now as before, by choosing a subsequence of $(x_n)_n$ we may assume WLOG that $(x_n)_n$ converges to a point $x$. We have $x \in X_n $ for every $n$ because $X_n$ is closed and $(x_j)_{j \geq n}$ is a sequence in $X_n$ converging to $x$ .So $x \in \cap_n X_n =Y$, so by continuity of $T$ we have $T(x)=y$ . So we have $Y \subset T(Y)$ .On the other hand, $z \in Y \implies (\forall n (T(z) \in T(X_n)=X_{n+1}\subset X_n)) \implies (\forall n (T(z) \in X_n))\implies T(z) \in Y$ .So we have $T(Y)\subset Y.$ ......................(8) FINALLY if $Y$ had more than one member then by (4) we have $D(T(Y)<D(Y)$ which is absurd because $T(Y)=Y$.And $Y\ne \phi$ so for some $y$ we have $\{y\}=Y=T(Y)=\{T(y)\}.$ QED.

Answer 3

If you want to keep it simple, using only elementary tools in analysis, you can just define the sequence $x_{n+1}=T(x_n)$, with $x_0$ arbitrary. First claim that the sequence is bounded (why?). Then prove that the sequence is Cauchy (you will need the previous step here). Then argue that the sequence converges since the space is also complete (why?). Finally, show that the limit is a fixed point for $T$ (this is an immediate consequence of the fact that $\{x_n\}$ is Cauchy).

As for the uniqueness, your proof is good.