# Independence of the uniform metric from the compatible metric for the codomain

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.
• 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. – Eric Wofsey Apr 5 at 15:24