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?

  • $\begingroup$ 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

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.

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