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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?

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  • $\begingroup$ Wht is your definition of “totally bounded”? $\endgroup$ – José Carlos Santos Oct 23 '19 at 10:40
  • $\begingroup$ @JoséCarlosSantos please see the Edit. I have added definition of totally bounded aswell as the result that I am using $\endgroup$ – Abhay Oct 23 '19 at 10:44
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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.

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  • $\begingroup$ I have used the result that $(X,d)$ is totally bounded $iff$ for every sequence we can get a Cauchy subsequence. $\endgroup$ – Abhay Oct 23 '19 at 10:45
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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.)

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