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.
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 $d_2$.