# Compactness of a metric space

If a metric space $$(X,d)$$ is compact then for every equivalent metric $$\sigma$$, $$(X,\sigma)$$ is complete. This is because, for any Cauchy sequence in $$(X,\sigma)$$ has a convergent subsequence due to fact $$(X,\sigma)$$ is a compact metric space, hence original sequence is convergent. My question is, does the converse also hold ?

In other words, let $$(X,d)$$ be a metric space such that for every equivalent metric $$\sigma$$ on $$X$$ is complete. Does this imply $$(X,d)$$ is compact?

• What definition of "equivalent metric" are you using? I understand it to mean the corresponding open sets are the same as the other. Sep 21, 2018 at 20:16
• A metric space (X,d) is said to be equivalent to the metric space (X,σ) iff topology induced by them are same i.e. if a subset U of X is d-open iff it is σ-open. Sep 22, 2018 at 2:59
• math.stackexchange.com/questions/2493757/… Jul 14, 2022 at 15:25

This is true but not very easy to prove.

Suppose $$X$$ is not compact. Without loss of generality assume that the original metric $$d$$ on $$X$$ is such that $$d(x,y)<1$$ for all $$x,y\in X.$$ There exists a decreasing sequence of non-empty closed sets $$\{C_{n}\}$$ whose intersection is empty. Let $$\rho (x,y)=\sum_{n=1}^{\infty }\frac{1}{% 2^{n}}d_{n}(x,y)$$ where $$d_{n}(x,y)=\left\vert d(x,C_{n})-d(y,C_{n})\right\vert +\min \{d(x,C_{n}),d(y,C_{n})\}d(x,y).$$ We claim that $$\rho$$ is a metric on $$X$$ which is equivalent to $$d$$ and that $$% (X,\rho )$$ is not complete. Note that $$d_{n}(x,y)\leq 2$$ for all $$x,y\in X.$$ If $$x$$ and $$y$$ $$\in C_{k}$$ then $$x$$ and $$y$$ $$\in C_{n}$$ for $$1\leq n\leq k$$ and hence $$\rho (x,y)\leq \sum_{n=k+1}^{\infty }\frac{2}{2^{n}}=% \frac{1}{2^{k}}$$. Thus, the diameter of $$C_{k}$$ in $$(X,\rho )$$ does not exceed $$\frac{1}{2^{k}}$$. Once we prove that $$\rho$$ is a metric equivalent to $$d$$ it follows that $$\rho$$ is not complete because $$\{C_{n}\}$$ is a decreasing sequence of non-empty closed sets whose intersection is empty.

Assuming (for the time being) that $$d_{n}$$ satisfies triangle inequality it follows easily that $$\rho$$ is a metric: if $$\rho (x,y)=0$$ then $$% d(x,C_{n})=d(y,C_{n})$$ for each $$n$$ and $$\min \{d(x,C_{n}),d(y,C_{n})\}d(x,y)=0$$ for each $$n$$. If $$d(x,y)\neq 0$$ it follows that $$d(x,C_{n})=d(y,C_{n})=0$$ for each $$n$$ which implies that $$x$$ and $$y$$ belong to each $$C_{n}$$ contradicting the hypothesis. Thus $$\rho$$\ is a metric. Also $$\rho (x_{j},x)\rightarrow 0$$ as $$j\rightarrow \infty$$ implies $$\left\vert d(x_{j},C_{n})-d(x,C_{n})\right\vert \rightarrow 0$$ and $$% \min \{d(x_{j},C_{n}),d(x,C_{n})\}d(x_{j},x)\rightarrow 0$$ as $$j\rightarrow \infty$$ for each $$n$$. There is at least one integer $$k$$ such that $$x\notin C_{k}$$ and we conclude that $$d(x_{j},x)\rightarrow 0$$. Conversely, suppose $$% d(x_{j},x)\rightarrow 0$$. Then $$d_{n}(x_{j},x)\rightarrow 0$$ for each $$n$$ and the series defining $$\rho$$ is uniformly convergent, so $$\rho (x_{j},x)\rightarrow 0$$. It remains only to show that $$d_{n}$$ satisfies triangle inequality for each $$n$$ . We have to show that $$\left\vert d(x,C_{n})-d(y,C_{n})\right\vert$$ $$+\min \{d(x,C_{n}),d(y,C_{n})\}d(x,y)$$

$$\leq \left\vert d(x,C_{n})-d(z,C_{n})\right\vert +\min \{d(x,C_{n}),d(z,C_{n})\}d(x,z)$$ $$+\left\vert d(z,C_{n})-d(y,C_{n})\right\vert +\min \{d(z,C_{n}),d(y,C_{n})\}d(z,y)$$ for all $$x,y,z.$$ Let $$% r_{1}=d(x,C_{n}),r_{2}=d(y,C_{n}),r_{3}=d(z,C_{n})$$. We consider six cases depending on the way the numbers $$r_{1},r_{2},r_{3}$$ are ordered. It turns out that the proof is easy when $$r_{1}$$ or $$r_{2}$$ is the smallest of the three. We give the proof for the case $$r_{3}\leq r_{1}\leq r_{2}$$. (The case $$r_{3}\leq r_{2}\leq r_{1}$$ is similar). We have to show that

$$r_{2}-r_{1}+r_{1}d(x,y)\leq r_{1}-r_{3}+r_{3}d(x,z)+r_{2}-r_{3}+r_{3}d(z,y)$$ which says $$r_{1}d(x,y)\leq 2r_{1}-2r_{3}+r_{3}d(x,z)+r_{3}d(z,y).$$ Since $$d$$ satisfies trangle inequality it suffices to show that $$% r_{1}d(x,z)+r_{1}d(z,y)\leq 2r_{1}-2r_{3}+r_{3}d(x,z)+r_{3}d(z,y).$$ But this last inequality is equivalent to $$(r_{1}-r_{3})[d(x,z)+d(z,y)]\leq 2r_{1}-2r_{3}.$$ This is true because $$d(x,z)+d(z,y)\leq 1+1=2$$.

• I believe I got this long ago from American Mathematical Monthly. Not my original proof. Sep 21, 2018 at 8:23

Nice question. Yes, it implies that $$(X,d)$$ is compact. In fact, suppose that $$(X,d)$$ is not compact. Then there is a sequence $$(F_n)_{n\in\mathbb N}$$ of closed subspaces of $$X$$ such that $$\bigcap_{n\in\mathbb N}F_n=\emptyset$$. For such a sequence, define the distance$$\sigma(x,y)=\sum_{n=1}^\infty\frac{\bigl\lvert d (x,F_n)-d(y,F_n)\bigr\rvert+\min\bigl\{d(x,F_n),d(y,F_n)\bigr\}\times d(x,y)}{2^n}.$$It can be proved (see Ryszard Engelking's General Topology, section 4.3) that:

• $$\sigma$$ is a distance equivalent to $$d$$;
• $$(X,\sigma)$$ is not complete.

Here is an easier proof by assuming that $$(X,d)$$ is locally compact and separable (the latter implies that it is also second countable).

Assume $$(X,d)$$ is not compact. Let $$S$$ be the one-point compactification of $$X$$. By the metrization theorems, $$S$$ is metrizable. Let $$\rho$$ be such a metrization of $$S$$ and let $$\sigma$$ be the restriction of $$\rho$$ to $$X$$. Then $$(X,\sigma)$$ is clearly not complete, since there exists a sequence in $$X$$ that converges (in $$(S,\rho)$$) to the unique point in $$S\setminus X$$. This completes the proof.

• The metrization theorem of Urysohn in the Wikipedia page is for second countable spaces. Sep 21, 2018 at 10:08
• One point compactificaton of a topological space X exists iff X is locally compact and hausdorff. Then how do you prove locally compactness of the metric space (X,d) whenever it is given every equivalent metric is complete. Sep 21, 2018 at 10:52
• Another comment regarding metrizability, it is well known that one point compactification of a locally compact hausdorff space is metrizable iff original space is second countable. Sep 21, 2018 at 10:58
• @UserS You are right, I have to assume second countable and locally compact. Should I delete my solution? Sep 21, 2018 at 13:40
• It's fine to leave it, so that future readers are aware of the pitfalls. Sep 21, 2018 at 15:53

Here is yet another proof.

Suppose $$(X,d)$$ is not compact. Since it is a metric space, it follows that there is a sequence $$\bar x = (x_n)_{n\in \mathbf N}$$ with no convergent subsequence. We will find an equivalent metric $$d'$$ which makes a subsequence of $$\bar x$$ a Cauchy sequence, so $$d'$$ is necessarily incomplete.

We may assume without loss of generality that $$\bar x = (x_n)_n$$ contains no $$d$$-Cauchy subsequence (otherwise, $$d=d'$$ is incomplete). It follows that there is some $$\varepsilon>0$$ such that for all $$n$$, there is some $$m>n$$ such that $$d(x_n,x_m)>\varepsilon$$. We may assume without loss of generality that $$d(x_n,x_m)>\varepsilon$$ for all $$n,m$$ (passing to a subsequence if necessary).

Now, for each $$n$$, let $$B_n$$ be the ball around $$x_n$$ of radius $$\varepsilon/2$$, and write $$X'$$ for $$X\setminus \bigcup_n B_n$$. Now, define a metric $$d'$$ by the following formula: $$d'(x,y) = \begin{cases} d(x,y) & x,y\in X' \\ d(x_n,y)+\varepsilon/n & x\in B_n, y\in X'\\ d(x,x_n)+\varepsilon/n & x\in X', y\in B_n\\ \varepsilon/\min(m,n) & x\in B_n, y\in B_m, n\neq m \\ d(x,y)/n & x,y\in B_n \end{cases}$$ It remains to check that:

1. $$d'$$ is a metric (the only nontrivial axiom is the triangle inequality, but it is easily checked by considering all possible cases),
2. $$d'$$ is equivalent to $$d$$ (this is very easy),
3. $$(x_n)_n$$ is Cauchy in $$d'$$ (this is trivial).