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This is my conjecture, but I guess I am missing the key idea for the proof (or my conjecture is wrong)

Let d and d' be two metrics on a compact space $X$ ($X$ is compact with respect to both metrics). If d' generates a finer topology than d on $X$, then for every $\epsilon >0$ there exists $\epsilon'>0$ such that for every $x \in X$ $$B_{d'}(x,\epsilon') \subseteq B_d(x,\epsilon)$$ holds. (So basically this $\epsilon'$ works for any $x \in X$)

My attempts are meaningless, I tried to use total boundedness, but it leads nowhere. So, can you please tell me if my conjecture is true or not, and if yes can you give me a proof or an idea for the proof?

Thanks for any help.

ThePortakal
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  • The conjecture seems to be true (and interesting): Any Cauchy net wrt to the one metric converges by completeness of compact metric spaces. But since the topologies agree it also converges wrt to the other metric. So it is especially Cauchy wrt to the other metric. Since nets capture all the information of topology and probably also of metrics, the spaces seem to be isomorphic as metric spaces... – C-star-W-star Jul 19 '14 at 05:39
  • The present title has a certain poetic incompleteness... Please modify it. – Did Jul 19 '14 at 06:49
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    Does one of the metrics generate the topology on $X$, under which $X$ is compact? – Henno Brandsma Jul 19 '14 at 06:54
  • And if so, which one? – C-star-W-star Jul 19 '14 at 10:06
  • BTW, the topologies generated by $d$ and $d'$ have to be the same: http://math.stackexchange.com/questions/800452/assume-that-textx-t-is-compact-and-hausdorff-prove-that-a-comparable-bu/800459#800459 – Najib Idrissi Jul 21 '14 at 08:05

1 Answers1

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Suppose that $(X,d')$ is compact (so $(X,d)$ is compact as well).

Given $\epsilon>0$, we have the cover $\left\{B_d(x,\frac{\varepsilon}{2}):x\in X\right\}$ of $X$ by $d'$-open sets. If $\epsilon'$ is a Lebesgue number for that cover, then $\epsilon'$ satisfies the condition you want.

Notice that the conjecture means that the identity $(X,d')\to (X,d)$ is uniformly continuous.

If we assume only $(X,d)$ to be compact (but not $(X,d')$), then this conjecture is not true, even if $(X,d')$ is complete.

For example, let $X=[-1,\infty)$ with the usual metric $d'$. Let $W$ be the Warsaw Circle (it's easier to understand it with the image). Let $f:[-1,\infty)\to W$ be a continuous, injective function such that $f(-1)=(0,1)$, $f(0)=(1,\sin 1)$ (so that $f$ covers the arc of $W$ in $[-1,0]$) and $f(x)=(e^{-x},\sin e^x)$ for $x>0$.

Consider the metric $d(x,y)=\Vert f(x)-f(y)\Vert_{\mathbb{R}^2}$ in $X$, so that $(X,d)$ is compact (because $f:(X,d)\to(W,\Vert\cdot\Vert_{\mathbb{R}^2})$ is a homeomorphism) and, since $f$ is continuous, then it follows easily that the usual metric (topology) $d'$ is finer than $d$, but the identity $(X,d')\to(X,d)$ is not uniformly continuous.

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Luiz Cordeiro
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