# Thread: Closed set

1. ## Closed set

I am working on this problem, and I am not 100% sure with my argument. Could someone help me verify it?

Let (X,d) be a metric space. Suppose that $$ is a sequence in X such that it has no convergent subsequence. Claim: $A=\{x_1,x_2,x_3,...\}$ is closed in X.

By assumption, for every sequence $$ in A, $$ does not converge. Otherwise, $$ has a convergent subsequence. This implies that A is closed since I'll show that a set A is closed in (X,d) iff for every sequence $$ in A if $$ converges then $lim p_n \in A$.

First direction: Suppose A is closed. Let $$ be a convergent sequence in A. Assume $lim p_n=p \notin A$, then $lim p_n \in X-A$. But X-A is open and $lim p_n \in X-A$ implies that there exists N such that if n>N, then $p_n \in X-A$. Contradicts to our assumption that $$ is a sequence in A.

Second direction: I'll show X-A is open. Let $$ be a sequence in X converging to $p \in X-A$. It suffices to show that there exists N such that if n>N, then $p_n \in X-A$. Assume $p_n \in X-A$ for only finitely many times, then $$ is a sequence in A. By hypothesis, $lim p_n \in A$. But $lim p_n=p \in X-A$. Contradiction. So, X-A is open, ie A is closed.

2. I think there is a technicality that you need to be careful of here.

Let $$ be the sequence $$. Then $$ is a sequence in $A$, but it is not a subsequence of $$.

Thus your statement following "by assumption" is not in fact by assumption, but must be proved.

3. THEOREM If $A$ is a subset of a metric space $(M,d)$ and $p$ is a limit point of $A$ then there is sequence of distinct points from $A$ that converges to $p$.
If you can prove that theorem the you have done this question.
i.e. The set has no limits points.

4. Originally Posted by DrSteve
I think there is a technicality that you need to be careful of here.

Let $$ be the sequence $$. Then $$ is a sequence in $A$, but it is not a subsequence of $$.

Thus your statement following "by assumption" is not in fact by assumption, but must be proved.
Thanks a lot for your correction, DrSteve. I should have seen that.
Thanks Plato. I think I can prove that statement. Could you verify my argument for me? Since $p$ is a limit point of $A$, every neighborhood of $p$ has nonempty intersection with $A-\{p\}$. So, there exists $p_n \in B(p,1/n) \bigcap A-\{p\}$. The sequence $$ converges to p. But how can I be sure that this is a sequence of distinct points. Also, I don't see how proving this means I'm done with the original question. Could you elaborate on that please?

5. Originally Posted by jackie
how can I be sure that this is a sequence of distinct points.
For each $n$ make sure that $\delta_n=\min\left\{\frac{1}{n},d(p,x_{n-1}\right)\}$.
Pick $p_n\in\mathcal{B}(p;\delta_{n})\setminus\{p\}$

6. From a PM.
You know that no subsequence of $$ converges.
So we have a contradiction.