# $(X,d)$ Complete and totally bounded $\iff$ $(X,d)$ is compact

Completeness and total boundedness $$\iff$$ compactness

$$(X,d)$$ Totally bounded means that $$\forall \epsilon >0 \; \exists n(\epsilon) \in N$$ and $$\exists x_1\ldots x_n \in X$$ such that $$X=\cup_{i=1}^{n} B_{\epsilon}(x_i)$$

Here I will repeatedly use the result : $$(X,d)$$ is totally bounded $$\iff$$ $$\forall (x_n)\in X\;\; \exists (x_{n_k})$$ which is Cauchy.

$$(\Rightarrow)$$ Let $$(X,d)$$ be complete and totally bounded metric space.

I will show that it is sequentially compact, thereby implying that it is compact.

Let $$(x_n)$$ be any sequence in $$X$$. I want to show that it has a convergent subsequence in $$X$$.

$$X$$ is totally bounded $$\rightarrow \; \exists (x_{n_k})$$ Cauchy subsequence of $$(x_n)$$

now, $$X$$ is also complete so that we have $$(x_{n_k})\to x_0$$ where $$x_0 \in X$$. So we have produced a convergent subsequence of $$(x_n)$$

Hence, it is sequentially compact and thus compact.

$$(\Leftarrow)$$ Now let $$(X,d)$$ be compact

Let $$(x_n)$$ be any sequence in $$X$$, then it has a convergent subsequence ($$X$$ is sequentially compact) and hence this subsequence is the required Cauchy subsequence. So $$X$$ becomes totally bounded.

Let $$(x_n)$$ be any Cauchy subsequence in $$X$$. Again by sequential compactness, it has a convergent subsequence (say it converges to $$x_0$$). So original sequence $$(x_n)$$ also converges to $$x_0$$. Hence it is also complete.

Is this correct?

• Wht is your definition of “totally bounded”? Oct 23, 2019 at 10:40
• @JoséCarlosSantos please see the Edit. I have added definition of totally bounded aswell as the result that I am using Oct 23, 2019 at 10:44

I don't understand your $$\Leftarrow$$ proof. In particular to prove that $$X$$ is totally bounded. A space $$X$$ is Totally bounded if and only if for every real number $$\varepsilon >0$$, there exists a finite collection of open balls in $$X$$ of radius $$\varepsilon$$ whose union contains $$X$$.

I would do the following. Consider the balls $$\mathcal B = \{B_\varepsilon(x) \mid x \in X\}$$. This is obviously an open cover of $$X$$. As $$X$$ is compact, we can extract a finite open subcover $$\overline{\mathcal B}$$ of $$\mathcal B$$. This proves that $$X$$ is totally bounded.

• I have used the result that $(X,d)$ is totally bounded $iff$ for every sequence we can get a Cauchy subsequence. Oct 23, 2019 at 10:45

I don't believe the result that $$X$$ is totally bounded $$\iff$$ for every sequence we can get a Cauchy subsequence is at all trivial.

Suppose $$X$$ is complete and totally bounded. Let $$x_n$$ be a sequence in $$X$$. Let $$\varepsilon>0$$ and choose a finite collection of open balls $$B_1,\ldots,B_j$$ with radius $$\varepsilon$$. Then as $$x_n$$ has infinitely many terms, some $$B_i$$ must contain infinitely many of the $$x_n$$. Hence the terms within that $$B_i$$, when ordered, form a Cauchy subsequence of $$x_n$$, and since $$X$$ is complete, it follows immediately that there is a convergent subsequence and hence $$X$$ is compact.

Conversely, suppose $$X$$ is compact. It is easy to see that $$X$$ is totally bounded, as the collection of open balls with radius $$1$$ centered at each point of $$X$$ is an open cover of $$X$$, but by compactness $$X$$ is contained within the union of finitely many of these. Now let $$x_n$$ be a sequence in $$X$$, then by sequential compactness $$x_n$$ has a convergent subsequence, and therefore $$X$$ is complete.

Note: It remains to be proved that compactness is equivalent to sequential compactness for metric spaces. (This is not true for arbitrary topological spaces.)

I'll show here that if $$X$$ is complete and totally bounded, then $$X$$ is compact. The other direction can be proved using contradiction.

Let $$(x^{(n)})_{n=1}^{\infty}$$ be a sequence in $$X$$. Since $$X$$ is totally bounded, it is contained in a finite number of balls of radius 1. Then there is at least one such ball that contains infinitely many terms of the sequence $$(x^{(n)})_{n=1}^{\infty}$$. Call it $$B(y^{(1)}, 1)$$. Let $$(x^{(n;1)})_{n=1}^{\infty}$$ be the subsequence of $$(x^{(n)})_{n=1}^{\infty}$$ that selects for each $$n$$ the $$n^{th}$$ occurrence in the ball $$B(y^{(1)}, 1)$$, of the elements in $$(x^{(n)})_{n=1}^{\infty}$$, in their original ordering. Hence $$(x^{(n;1)})_{n=1}^{\infty} \subseteq B(y^{(1)}, 1)$$. Since $$B(y^{(1)}, 1) \subseteq X$$, $$B(y^{(1)}, 1)$$ is totally bounded and thus can be covered by a finite number of balls of radius 1/2. Like before, we can extract a subsequence $$(x^{(n;2)})_{n=1}^{\infty}$$ of $$(x^{(n;1)})_{n=1}^{\infty}$$, that is contained in a ball $$B(y^{(2)}, 1/2)$$. Keep extracting subsequences using radius of $$1/k, k=1,2,...$$ We end up with a sequence $$(x^{(n;1)})_{n=1}^{\infty}, (x^{(n;2)})_{n=1}^{\infty} ...$$ of subsequences, such that for each $$j$$, the elements of the sequence $$x^{(n;j)})_{n=1}^{\infty}$$ are contained in a single ball of radius $$1/j$$, and also that each sequence $$x^{(n;j+1)})_{n=1}^{\infty}$$ is a subsequence of the previous one $$x^{(n;j)})_{n=1}^{\infty}$$.

We claim that the sequence $$(y^{(n)})_{n=1}^{\infty}$$ has to be Cauchy. Suppose that $$1 \leq r \leq s$$. By our construction, $$x^{(1;s)} \in B(y^{(s)}, 1/s) \cap B(y^{(r)}, 1/r)$$. ie, $$d(y^{(s)}, x^{(1;s)}) < 1/s$$ and $$d(y^{(r)}, x^{(1;s)}) < 1/r$$. By the Triangle inequality, $$d(y^{(s)}, y^{(r)}) < 1/r + 1/s$$, which can be made arbitrarily small by taking $$r$$ and $$s$$ sufficiently large.

Now, Consider the diagonal sequence $$(x^{(n;n)})_{n=1}^{\infty}$$. Let $$1 \leq r \leq s$$, we have $$d(x^{(r;r)}, x^{(s;s)}) \leq d(x^{(r;r)}, y^{(r)}) + d(y^{(r)}, y^{(s)}) + d(y^{(s)} ,x^{(s;s)}) < 1/r + 1/s + o(r, s)$$. Hence $$(x^{(n;n)})_{n=1}^{\infty}$$ is Cauchy.

Finally, since $$X$$ is complete, the sequence $$(x^{(n;n)})_{n=1}^{\infty}$$ converges. By definition, this implies that $$X$$ is compact.