Let $X$ be a compact metrizable space and $Y$ a metrizable space. Denote by $C(X,Y)$ the space of continuous functions from $X$ to $Y$ with the topology induced by the uniform metric $$d_u(f,g)=sup_{x\in X}d(f(x),g(x)).$$ where $d$ is a compatible metric for $Y$. At pp. $24$ "Classical Descriptive Set Theory", Kechris states:

A simple compactness argument shows that this topology is indipendente of the choice of $d$.

The problem is that it is not clear to me at all how such a simple argument should work, at least looking at definitions.

My attempt (while the below answer was edited): Assume $d_1$, $d_2$ are compatible metrics on $Y$. Let $$B_{d_1}(f,\epsilon)=\{g\in C\mid sup_{x\in X} d_1(f(x),g(x))<\epsilon\} \supseteq\{g\in C\mid d_1(f(x),g(x))<\epsilon'\, \text{for every }x\in X\}=B(f,\epsilon')\supseteq\{g\in C\mid sup_{x\in X} d_1(f(x),g(x))<\frac{\epsilon}{2}\}=B_{d_1}(f,\frac{\epsilon}{2}).$$ if we choose $\frac{\epsilon}{2}<\epsilon'<\epsilon$. By compactness, as the family of open balls with radius $\frac{\epsilon}{2}$ is an open cover of $f(X)$, we can assume the condition holds at points $f(x_1),\dots,f(x_n)$. As $d_1$, $d_2$ are compatible, there is $\delta $ s.t. $B_{d_2}(f(x_i),\delta)\subseteq B_{d_1}(f(x_i),\frac{\epsilon}{2})$ for every $i=1,\dots,n$. Does this suffice to prove that for every $\epsilon$ there is $\delta$ s.t. $B_{d_2}(f,\delta)\subseteq B_{d_1}(f,\epsilon)$? If this is the case, the proof applies switching $d_1$, $d_2$.


Let $d'$ be another metric on $Y$ and $d_u'$ be the associated uniform metric. Fix $f\in C(X,Y)$ and $\epsilon>0$. For each $x\in X$, pick $\epsilon_x>0$ such that $d'(f(x),y)<\epsilon_x$ implies $d(f(x),y)<\epsilon$. Also pick a neighborhood $U_x$ of $x$ such that $d'(f(x),f(y))<\epsilon_x/2$ for all $y\in U_x$. By compactness, $X$ is covered by finitely many of these neighborhoods $U_{x_1},\dots,U_{x_n}$. Let $\epsilon'=\min(\epsilon_{x_1},\dots,\epsilon_{x_n})$.

Now suppose $g\in C(X,Y)$ is such that $d_u'(f,g)<\epsilon'/2$ and let $x\in X$. There is some $i$ such that $x\in U_{x_i}$. Since $x\in U_{x_i}$, $d'(f(x),f(x_i))<\epsilon_{x_i}/2<\epsilon_{x_i}$ and so $d(f(x),f(x_i))<\epsilon$. Also, $$d'(g(x),f(x_i))\leq d'(g(x),f(x))+d'(f(x),f(x_i))<\epsilon'/2+\epsilon_{x_i}/2\leq \epsilon_{x_i}$$ so also $d(g(x),f(x_i))<\epsilon$. Thus $$d(f(x),g(x))\leq d(f(x),f(x_i))+d(g(x),f(x_i))<2\epsilon.$$ That is, $d_u'(f,g)<\epsilon'/2$ implies $d_u(f,g)\leq 2\epsilon$. Since $f$ and $\epsilon>0$ were arbitrary, this means the $d_u'$-topology is finer than the $d_u$ topology, and swapping the roles of $d$ and $d'$ we conclude the two topologies are the same.

Alternatively, you can show that the topology induced by $d_u$ is the compact-open topology, that is the topology generated by sets of the form $\{f:f(K)\subseteq U)$ where $K\subseteq X$ is compact and $U\subseteq Y$ is open. This manifestly does not depend on the choice of a metric on $Y$. The hard part of the proof is to show a $d_u$-ball around $f$ contains a compact-open neighborhood of $f$. The idea is that you can cover $X$ by finitely many closed balls $K_i$ such that $f(K_i)$ has small diameter, and let $U_i$ be a small open neighborhood of $f(K_i)$. Then if $g$ is such that $g(K_i)\subseteq U_i$ for all $i$, $d_u(f,g)$ must be small since each $U_i$ has small diameter.

  • $\begingroup$ I read your very plain proof and everything is clear. While you wrote your answer, I was editing an attempt. I have only one request: could you check if the above argument is correct, please? Thank you $\endgroup$ – LBJFS Apr 5 at 8:21
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    $\begingroup$ Well, it's quite obviously not a complete proof since you haven't explained why $B_{d_2}(f,\delta)\subseteq B_{d_1}(f,\epsilon)$. And I don't think the argument can work: you need to use the continuity of $f$ somewhere. $\endgroup$ – Eric Wofsey Apr 5 at 15:24
  • $\begingroup$ Ok, I thank you for your explanations :) $\endgroup$ – LBJFS Apr 5 at 15:28

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