(verification) Set of Cluster points of a sequence $x_k$ is closed.

Question

Claim

Set of Cluster(Accumulation) points of a sequence $x_k$ is closed.

Proof

A set is closed if and only if its accumulation(limit) point in that set.

So we can change the claim into "set of cluster points of a sequence $x_k$ holds the limit points (of the set of cluster points of a sequence)".

Think about what has to appen for a point $x\in \overline{ {x_k}}$ I mean, if $x$ is in the closure of the sets of cluster points, it is clear that $x$ is also in the closure of the sequence. (You should prove it) — jJjjJ, Jun 01 '17 at 11:02
You can change the claim into: "set of clusterpoints of sequence holds its limitpoints." But be aware that here "its" does not refer to the sequence itself but to the set of clusterpoints of the sequence. Also it can be convenient to look at Wikipedia about the difference between limitpoints of sets and the clusterpoints of sequences. — drhab, Jun 01 '17 at 11:45
@drhab could you define the limit point of sequence? Wikipedia doesn't provide nice and crisp definition of it. — Daschin, Jun 01 '17 at 12:21
For me $x$ is a limit point of sequence $(x_n)n$ iff there is a subsequence $(x{n_k})$ that converges to $x$. Also have a look here. — drhab, Jun 01 '17 at 12:23
@drhab what is $(x_n)_n$ denoting? two ns looks unfamiliar to me. — Daschin, Jun 01 '17 at 12:26
It denotes a sequence and abbreviates $(x_n)_{n=1}^{\infty}$. $(x_n)$ will do too. — drhab, Jun 01 '17 at 12:28
@drhab I see. That denotation was the reason that I couldn't understand the body of wiipedia page — Daschin, Jun 01 '17 at 12:36

score 1 · Answer 1 · answered Jun 01 '17 at 11:08

Let $S$ denote the set of elements that are not clusterpoints of sequence $(x_n)_n$ and let $s\in S$.

By definition an open set $U_s$ exists such that $\{x_n\in U_s\mid n\in\mathbb N\}$ is finite.

But actually this shows that every element of $U_s$ is not a clusterpoint of the sequence.

So we have $U_s\subseteq S$.

Now observe that $S=\bigcup_{s\in S}U_s$ which - as union of open sets - is open itself.

Then its complement, i.e. the set of clusterpoints of sequence $(x_n)_n$ is closed.

change the OP. please verify when you have a chance. – Daschin Jun 01 '17 at 11:34 — Daschin, Jun 01 '17 at 11:34

score 0 · Answer 2 · answered Jun 01 '17 at 11:57

Denote the set of cluster points of a set $E$ by $E'$.

If $\{x_n\}''$ is empty, then $\{x_n\}'$ is closed.

Suppose $x \in \{x_n\}''$. Let $U$ be an open set containing $x$.

By definition, $U$ contains infinitely many points of $\{x_n\}'$. Let $y$ be one such point.

Then $U$ is an open set containing $y \in \{x_n\}'$ so again by definition $U$ contains infinitely many points of $\{x_n\}$.

To recap, any open set containing $x$ contains infinitely many points of $\{x_n\}$ so that $x \in \{x_n\}'$.

2 Answers2