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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 ?

Martin Argerami
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James Well
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2 Answers2

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$\newcommand{\Reals}{\mathbf{R}}$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.

Andrew D. Hwang
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  • in your last example Does both d and d' induce same topology on $X$? – Meet Patel May 16 '23 at 15:47
  • Can you demonstrate a totally bounded metric on $\mathbb R$? – Atom Mar 18 '24 at 20:29
  • @Atom Up to isometry, the absolute value metric on a bounded open interval "is" a totally-bounded metric on the reals. – Andrew D. Hwang Mar 24 '24 at 01:53
  • @AndrewD.Hwang I agree that in this way, you can endow $\mathbb R$ with a totally bounded metric that will topologically be equivalent to the usual Euclidean metric on $\mathbb R$. But how do you ensure uniform equivalence? – Atom Mar 26 '24 at 09:57
  • @AndrewD.Hwang I understand that you are proposing to lift the (usual Euclidean) metric on, say, $(-1, 1)$ to all of $\mathbb R$ by a bijection $f\colon \mathbb R\to (-1, 1)$ by defining $d'(x, y) := d(f(x), f(y))$. Is that correct? [1/2] – Atom Mar 26 '24 at 10:26
  • If so, then it's easy to show that $d'$ is uniformly equiv to $d$ $\iff$ $f$ and $f^{-1}$ are Lipschitz continuous (with the usual metrics on the domains and codomains). But uniform continuity of $f^{-1}$ (due to its Lipschitz-ness) implies that the image of $f^{-1}$, which is $\mathbb R$, be bounded, a contradiction. This ensures that the above method won't work for producing a uniformly equivalent metric. [2/2] – Atom Mar 26 '24 at 10:45
  • @Atom FWIW, I'm not sure about your "if and only if Lipschitz" claim (quotes indicating my paraphrase, not your verbatim), because of bijections like $x \mapsto x^3$. But that's a weedy point; the gist of your comments looks sound on first read. <> Does there seem to be an error in this answer, or are you asking for amplification and/or about a related issue not addressed here? If either of the latter, asking a new question with a link to this one may be the best path forward. – Andrew D. Hwang Mar 27 '24 at 12:36
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    @AndrewD.Hwang My bad. I don't know what got into my head, making me think that you were demanding a uniformly equivalent metric. In fact, as you clearly write in your answer, you intended the metric to be only (topologically) equivalent. In that case, $f$ and $f^{-1}$ just need to be continuous (as opposed to be Lipschitz). You answer makes complete sense. A big +1! – Atom Mar 29 '24 at 03:26
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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.

Martin Argerami
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