1
$\begingroup$

Let $(X,d)$ be a metric space, and let $(x_n)_{n=1}^\infty$ be a sequence in $X$ with limit $x_0$. Show that the subset $\{x_0, x_1, x_2, \dots\}$ of $X$ is compact.

This is a book problem from A Taste of Topology that I am trying to understand using the "every open cover has a finite subcover" method to prove.

My attempt at a solution:

Since $\{x_n\}$ is a convergent sequence, it is bounded. Hence, we can say that $\{x_k : k\ge n\}$ is finite. Thus, the subset $\{x_0, x_1, x_2, \dots\}$ is also finite. Consequently, we can generate a finite subcover for every open cover for the subset $\{x_0, x_1, x_2, \dots\}$. Therefore, the subset $\{x_0, x_1, x_2, \dots\}$ of $X$ is compact.

Is my logic sound and also if so, how can I show how to generate such a finite subcover?

$\endgroup$
  • $\begingroup$ Your proof as it stands has some issues. As an example for $\{x_n\}$, consider let $x_k=1/k$ for $k\geq 1$ and $x_0=0$ which converges to 0. Your ambiguous language and notation seems to conclude that there are finitely many $x_n$'s? Your intuition is right. Boundedness will play a strong role here (as a counter example, consider $x_k=k$, with no limit). You somehow need to show that any cover will heavily overlap around $x_0$ making all but finitely many of the covers redundant. Here is a very strong hint: $x_0$ must belong to one of the covers! $\endgroup$ – Alex R. Oct 5 '13 at 19:51
  • $\begingroup$ It is almost obvious. You only need to work with the definition of convergence in terms of open sets. $\endgroup$ – user56706 Oct 5 '13 at 19:51
3
$\begingroup$

No, the set $\{x_0, x_1, \ldots\}$ is not finite. Hint: For any open cover one of those open sets will contain $x_0$. Now use the definition of convergence.

| cite | improve this answer | |
$\endgroup$

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.