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?

  • $\begingroup$ What definition of "equivalent metric" are you using? I understand it to mean the corresponding open sets are the same as the other. $\endgroup$ – TurlocTheRed Sep 21 '18 at 20:16
  • $\begingroup$ 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. $\endgroup$ – Mathlover Sep 22 '18 at 2:59

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 $d(x,z)+d(z,y)\leq 1+1=2$.

  • $\begingroup$ I believe I got this long ago from American Mathematical Monthly. Not my original proof. $\endgroup$ – Kavi Rama Murthy Sep 21 '18 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.

  • 3
    $\begingroup$ The metrization theorem of Urysohn in the Wikipedia page is for second countable spaces. $\endgroup$ – Kavi Rama Murthy Sep 21 '18 at 10:08
  • $\begingroup$ 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. $\endgroup$ – Mathlover Sep 21 '18 at 10:52
  • $\begingroup$ 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. $\endgroup$ – Mathlover Sep 21 '18 at 10:58
  • $\begingroup$ @UserS You are right, I have to assume second countable and locally compact. Should I delete my solution? $\endgroup$ – Ali Khezeli Sep 21 '18 at 13:40
  • 2
    $\begingroup$ It's fine to leave it, so that future readers are aware of the pitfalls. $\endgroup$ – user21820 Sep 21 '18 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).

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