Uniform continuity preserved with equivalent metrics? I am told that any two metrics that equip a space with the same topology yield the same uniformly continuous functions. Surely this is not true ?
The reason I ask is because in one of my exams I'm expected to prove that some function is uniformly continuous on $[0,1]^2$ and when trying to do so I found myself stuck not knowing exactly relative to which metric. I asked the examinor and he said "any one of them that equips it with its natural topology" but I'm pretty sure that doesn't work out. Are there two metrics that equip this space with its natural product topology and a function which is uniformly continuous wrt one and not the other ?
 A: $\newcommand{\Reals}{\mathbf{R}}$If $X$ is compact, the statement is true, by Martin Argerami's observation: If $(X, d)$ and $(X, d')$ are equivalent, the identity map of $X$ is a homeomorphism, so $f$ is uniformly continuous on $(X, d)$ if and only if $f$ is continuous on $(X, d)$, if and only if $f$ is continuous on $(X, d')$, if and only if $f$ is uniformly continuous on $(X, d')$.
The claim is generally false for non-compact spaces, however: Let
$$
X = S^{1} \setminus\{(-1, 0)\} = \{e^{i\theta} : -\pi < \theta < \pi\},
$$
and define metrics by
$$
d(e^{i\theta_{1}}, e^{i\theta_{2}}) = |\theta_{2} - \theta_{1}|,\qquad
d'(e^{i\theta_{1}}, e^{i\theta_{2}}) = |e^{i\theta_{1}} - e^{i\theta_{2}}|.
$$
Each of $(X, d)$ and $X, d')$ is diffeomorphic to the open interval $(-\pi, \pi)$, but the function $\theta:X \to (-\pi, \pi)$ is uniformly continuous on $(X, d)$ (where it is essentially the identity function on the open interval $(-\pi, \pi)$) and not uniformly continuous on $(X, d')$ (for every $\delta > 0$, the points $x = e^{i(-\pi + \delta/2)}$ and $y = e^{i(\pi - \delta/2)}$ are $\delta$-close with respect to $d'$, but $|\theta(x) - \theta(y)| \approx 2\pi$).
Alternatively, let $X = (-\frac{\pi}{2}, \frac{\pi}{2})$, and let $d$ be the Euclidean metric and $d'(x, y) = |\tan(x) - \tan(y)|$. The tangent function is obviously uniformly continuous on $(X, d')$, but not on $(X, d)$.
To express the same idea from a different point of view, if $X = \Reals$, $d$ is the Euclidean metric, and $d'$ is any equivalent totally bounded metric, then every unbounded, $d$-uniformly continuous function (such as $f(x) = x$) fails to be uniformly continuous on $(X, d')$, since a uniformly-continuous function on a totally-bounded metric space is bounded.
A: I'm not sure about the general answer to your question. But in your concrete example $[0,1]^2$ is compact, so whatever metric you choose all you need to show is continuity: a continuous function on a compact metric space is uniformly continuous. 
