Let $(x_k)$ be a convergent sequence in $\mathbb{R}$ such that $(x_k) \to x$. Let $A = \{x_1, x_2, \ldots\} \cup \{x\}$.

It's quite easy to show that $A$ is compact by showing that every open cover has a finite subcover. And since $\mathbb{R}$ is a metric space, it follows that $A$ is sequentially compact.

But I need to show directly that $A$ is sequentially compact. So how can I show that every sequence in $A$ has a subsequence that converges in $A$?


1 Answer 1


Hint: choose $y_n$ a sequence in $A$. If $y_n$ has only finitely many terms you are done. Otherwise keep taking terms of $y_k$ that correspond to higher and higher indexed terms of $x_n$ to show that $y_k$ has a subsequence converging to $x$.


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.