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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”? $\endgroup$ Commented Oct 23, 2019 at 10:40
  • $\begingroup$ @JoséCarlosSantos please see the Edit. I have added definition of totally bounded aswell as the result that I am using $\endgroup$
    – chesslad
    Commented Oct 23, 2019 at 10:44

3 Answers 3

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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.

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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. $\endgroup$
    – chesslad
    Commented Oct 23, 2019 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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