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 $\displaystyle <x_n>$ is a sequence in X such that it has no convergent subsequence. Claim: $\displaystyle A=\{x_1,x_2,x_3,...\}$ is closed in X.

By assumption, for every sequence $\displaystyle <p_n>$ in A, $\displaystyle <p_n>$ does not converge. Otherwise, $\displaystyle <x_n>$ 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 $\displaystyle <p_n>$ in A if $\displaystyle <p_n>$ converges then $\displaystyle lim p_n \in A$.

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

Second direction: I'll show X-A is open. Let $\displaystyle <p_n>$ be a sequence in X converging to $\displaystyle p \in X-A$. It suffices to show that there exists N such that if n>N, then $\displaystyle p_n \in X-A$. Assume $\displaystyle p_n \in X-A$ for only finitely many times, then $\displaystyle <p_n>$ is a sequence in A. By hypothesis, $\displaystyle lim p_n \in A$. But $\displaystyle lim p_n=p \in X-A$. 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 $\displaystyle <p_n>$ be the sequence $\displaystyle <x_2,x_1,x_4,x_3,...>$. Then $\displaystyle <p_n>$ is a sequence in $\displaystyle A$, but it is not a subsequence of $\displaystyle <x_n>$.

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

3. THEOREM If $\displaystyle A$ is a subset of a metric space $\displaystyle (M,d)$ and $\displaystyle p$ is a limit point of $\displaystyle A$ then there is sequence of distinct points from $\displaystyle A$ that converges to $\displaystyle p$.
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 $\displaystyle <p_n>$ be the sequence $\displaystyle <x_2,x_1,x_4,x_3,...>$. Then $\displaystyle <p_n>$ is a sequence in $\displaystyle A$, but it is not a subsequence of $\displaystyle <x_n>$.

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 $\displaystyle p$ is a limit point of $\displaystyle A$, every neighborhood of $\displaystyle p$ has nonempty intersection with $\displaystyle A-\{p\}$. So, there exists $\displaystyle p_n \in B(p,1/n) \bigcap A-\{p\}$. The sequence $\displaystyle <p_n>$ 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 $\displaystyle n$ make sure that $\displaystyle \delta_n=\min\left\{\frac{1}{n},d(p,x_{n-1}\right)\}$.
Pick $\displaystyle p_n\in\mathcal{B}(p;\delta_{n})\setminus\{p\}$

6. From a PM.
You know that no subsequence of $\displaystyle <x_n>$ converges.