Compactness of a metric space

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

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.