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?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.