Suppose that for every sequence $s_n$ in A, the limit of $s_n$ is in A. Show that A must be closed.

Question

Let $A \subset \mathbb{R}$.

Suppose that for every sequence $s_n$ in A, the limit of $s_n$ is in A. Show that A must be closed.

My Attempt:

Let $s_n$ be a sequence in A and $s=lim(s_n)$. We know that if $s_n,s\in A$ and $s_n \to s $ then s is a cluster point of A. We also know that since every $s \in A$, then A contains all it cluster points, thus implying A is closed.

Is this correct?

Answer 1

That is not adequate. You've shown (actually, you've just stated, but I assume this has already been covered in your course) that

$$(s = \lim s_n \text{ with } \{s_n\} \subseteq A) \implies (s\text{ is a cluster point of }A)$$

But your conclusion requires $$(s\text{ is a cluster point of }A)\implies (s = \lim s_n \text{ with } \{s_n\} \subseteq A)$$

Until you've proven that implication, your proof is incomplete.