# Uniform convergence not a topological property?

Suppose I have a sequence of functions $f_n: X \to Y$ between metric spaces that converges uniformly to $f$. Then if I change the metric on $Y$ to a Lipschitz equivalent metric, the sequence of functions will still converge uniformly to $f$.

However, I am told that if I change the metric on $Y$ in such a way that does not disturb the induced topology, then $f_n$ may no longer converge uniformly. My question is: is there a simple example of a sequence of functions from $X$ to two homeomorphic metric spaces $Y$, $Y'$ converging uniformly in one but not the other?

EDIT: question modified slightly in response to comments.

• The different metric must be on $Y$, not on $X$, because the metric (not even the topology, actually) of $X$ never really comes into play. You could pick a sequence of functions $f_n:A\to (Y,d)$, where $A$ is just a set, and define uniform convergence in the same way. – user228113 Jan 13 '17 at 12:05
• No, the domain $X$ need not even be a topological space to speak of uniform convergence of a sequence (filter, net) of functions $X\to Y$. For that, only $$\sup \{ d(f_n(x), f(x)) : x \in X\}$$ is relevant, and that only involves the metric (or more generally the uniform structure) on the codomain. – Daniel Fischer Jan 13 '17 at 12:15
• Thanks for the reply --- I just realised I was being silly (indeed $X$ can be any set) and deleted my comment. – gj255 Jan 13 '17 at 12:16
• The topology (more precisely, the metric, or more broadly, the uniformity) on $X$ will matter when you want to talk about uniform continuity of a function. – tomasz Jan 13 '17 at 12:42

$Y$ must not be compact for this to be possible (because compact spaces have unique uniform structure).
You can simply construct an artificial example. For instance, let $X={\bf N}$ and $Y={\bf N}\times [0,1]$, while $Y'$ is $Y$ with metric on $\{n\}\times [0,1]$ scaled by $1/n$ (so that the diameter is $1/n$, you can put the distance between components in any way you want, for example put $2$ whenever two points are in different components). Clearly, $Y'$ is homeomorphic to $Y$ (even by a locally bi-Lipschitz function), but if you take sequence of functions $(f_m)$ with $$f_m(n)=\begin{cases}(n,1) &\textrm{if }m>n\\ (n,0) &\textrm{otherwise}\end{cases}$$ you will have $f_m$ converge uniformly as sequence of functions into $Y'$, but not so into $Y$. You can replace $[0,1]$ here with any nontrivial metric space.
For a different example, take for $Y$ the interval $(0,1]$ with the standard metric, and for $Y'$ the same interval with the metric scaled by a singular function, e.g. for $y_1<y_2$ put $d(y_1,y_2)=y_1^{-1}-y_2^{-1}$, and put $$f_m(n)=\begin{cases}\frac{1}{n} &\textrm{if }m>n\\ \frac{1}{n+1} &\textrm{otherwise}\end{cases}$$ This sequence will be uniformly convergent in the standard metric, but not in the scaled metric.
• Thanks a lot for the answer. For the first function you defined, aren't the $f_m$ converging to $f: n \mapsto (n,1)$? – gj255 Jan 13 '17 at 13:33
Consider $f_n(x)=x+\frac1n$, from $X=(0,1)$ to $Y=\Bbb R$, with the usual metrics. This converges uniformly to $f(x)=x$.
Now, change the metric of the codomain to $d'(x,y)=\frac1x-\frac1y$. That is, let $Y'$ be $(\Bbb R,d')$. Then $f_n$, as a function from $X$ to $Y'$ no longer converges uniformly to $f$.