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Definition: Let $f_n:X\to Y$ be a sequence of functions from a set $X$ to the metric space $Y$. Let $d$ be the metric for $Y$. The sequence $(f_n)$ d-converges uniformly to the function $f:X\to Y$ if given $\varepsilon>0$, there exists an integer $N$ such that $d(f_n(x),f(x))<\varepsilon$ for all $n\geq N$ and all $x\in X$.

I wonder if this definition depends on the metric $d$. More explicitly: consider $d_1,d_2$ metrics on $Y$ such that the corresponding induced topologies are the same, and let $f:X \to Y$ a function.

Is the following statement true?: If $f_n:X\to Y$ is a sequence of functions that $d_1$-converges uniformly to $f$, then it also $d_2$-converges to $f$.

If the answer is no, it would be quite interesting to see a counter-example.

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    $\begingroup$ Certainly it remains true for all strongly equivalent metrics to $d$: en.m.wikipedia.org/wiki/Equivalence_of_metrics $\endgroup$ – Alex R. Sep 8 '14 at 4:55
  • $\begingroup$ Can you think of a pair of metrics that are topologically, but not strongly equivalent? $\endgroup$ – Alex R. Sep 8 '14 at 4:56
  • $\begingroup$ If $d$ is a metric then $\frac{d}{1+d}$ is a topologically equivalent one but generally they are not strongly equivalent. For example the usual metric in $\mathbb{R}$, $|x-y|$ is not bounded and its standard bounded version $\frac{|x-y|}{1+|x-y|}$ is bounded, so they cannot be strongly equivalent. $\endgroup$ – Chilote Sep 8 '14 at 5:03
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No, it's false.

Let $Y = (0,+\infty)$ with the metrics $d_1(x,y) = |y - x|$ and $d_2(x,y) = |\log y - \log x|$. The two metrics clearly define the same topology, since the second is isometric to $\mathbf{R}$ with the usual distance, and the exponential function is a homeomorphism of $\mathbf{R}$ onto $(0,+\infty)$.

Also let $X = (0, + \infty)$. Now define $f_n(x) = x + 1/n$. The sequence converges uniformly to $f(x) = x$ for $d_1$. However, $\sup_{x > 0} d_2(f_n(x),f(x)) = +\infty$, so $f_n$ doesn't converge uniformly for $d_2$.

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  • $\begingroup$ ...and hence $d_1$ and $d_2$ cannot be strongly equivalent. $\endgroup$ – Chilote Sep 8 '14 at 6:13
  • $\begingroup$ Yes. But your example of $d/(1 + d)$ wouldn't work. In fact, if it's possible to simply calculate the second distance from the first, and the two define the same topology, then uniform convergence in one is equivalent to uniform convergence in the other. $\endgroup$ – Dave Sep 8 '14 at 6:20

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