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Let $(X,\rho)$ be a non-empty compact metric space. I've learnt enough over the last week to know that that means it is complete, totally bounded and sequentially compact, and Cantor intersection works too. I have also learnt the Baire Category theorem, and the Arzela-Ascoli theorem. The following exercise is from Royden:

If $T:X\to X$ satisfies $\rho(T(u),T(v))\lt\rho(u,v),\,\forall u,v\in X$ then there exists a unique fixed point.

Of course I cannot use the Banach contraction theorem nor its core ideas, as although $T$ "contracts", the factor by which it contracts is arbitrarily close to $1$, and variable. Perhaps it is better to say that $T$ does not expand - I'm not sure what the terminology is. Maybe Lipschitz-$1$?

My ideas:

  1. $T$ is continuous, and $T(X)$ is compact and thus also a complete subspace - necessarily $T(X)$ is a closed subset of $X$, and inductively $T^n(X)$ is closed for all natural $n$. We then have that $\{T^n(X)\}$ is a "contracting" sequence of closed non-empty sets, and compactness allows us to utilise Cantor's Intersection Theorem (one version thereof) to say that $A=\bigcap_{n\in\Bbb N}T^n(X)\neq\varnothing$. If this intersection were a singleton, we would be done as that would imply the singular element was a fixed point, but as $T$ does not forcefully contract, we only have that $\operatorname{diam}T^{n+1}(X)\lt\operatorname{diam}T^n(X)$ but we do not have that the diameters tend to zero. Perhaps I can argue that $T(A)=A$ and if $A$ has non-zero diameter then this is a contradiction as no real number is less than itself, but it feels like a logical mistake since we can have monotonically decreasing sequences which converge to numbers other than zero.
  2. I can show (can’t post my derivation right now but I will do so later: it involves the sequential compactness of $X$ and applying that to $T^(x_0)$ for an arbitrary $x_0\in X$) that there exists at least one point $x$ which is recurrent in the sense that there is an increasing sequence of integers $m_k$ such that for all $\epsilon\gt0$ there is a $K$ such that $k\gt K\implies\rho(x,T^{m_k}(x))\lt\epsilon$. However, even though $T$ is continuous, this is not strong enough to imply $x=\lim_{n\to\infty}T^n(x)\therefore T(x)=x$, since $m_k$ very likely has gaps in the sequence and so the sequence $T^{m_k+1}(x)$ is not the same (necessarily) as $T^{m_k}(x)$.
  3. I can cover $T(X)$ with finitely many balls of fixed radius and pullback $T$ to obtain a cover of $X$ with finitely many balls of that radius or greater, and perhaps apply pigeonholing to say that at least one ball has a subset invariant under $T$, and then keep using this logic to claim the existence of a fixed point - however, I do not think this pigeonholing idea will work as there's no telling how much $T$ might "scatter" the balls.

Any hints? Are any of the above ideas along the right lines? Many thanks.

FShrike
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  • @ArcticChar It does... so much simpler than I thought. Thank you – FShrike Jan 18 '22 at 15:12
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    Another useful post is here : one of the answers completes the first approach you make. The point is that $A$ is compact (closed subset of compact set) and the function $\rho(\cdot,\cdot) : K^2 \to \mathbb R$ is continuous, so you can find $a_1,a_2 \in A$ such that $\rho(a_1,a_2)$ is the diameter of $A$. Then the argument shows that $a_1=a_2$ so that $A$ is a singleton. – Sarvesh Ravichandran Iyer Jan 18 '22 at 15:15
  • @SarveshRavichandranIyer Thank you for letting me know my efforts weren’t fruitless! – FShrike Jan 18 '22 at 15:21
  • @FShrike Thanks, it is perhaps what you deserve for adding an excellent amount of context. – Sarvesh Ravichandran Iyer Jan 18 '22 at 15:22

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