(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Let {x_{n}} be a sequence of real numbers. Let E denote the set of all numbers z that have the property that there exists a subsequence {x_{nk}} convergent to z. Show that E is closed.

2. Relevant equations

A closed set must contain all of its accumulation points.

Sets with no accumulation points are closed.

The Bolzano-Weierstrass theorem: every bounded sequence has a convergent subsequence.

3. The attempt at a solution

What I've done for this problem is I've split it up into several cases.

Case 1

{x_{n}} is convergent to some number z. Then all subsequences converge to that number z. Thus, E = {z}. Since E is a finite set (consisting of one isolated point), it is closed.

Case 2

{x_{n}} is unbounded and divergent. Then, by the Bolzano-Weierstrass theorem, there are no convergent subsequences. Thus, E=∅, which is both open and closed.

Case 3

{x_{n}} is bounded and divergent. Then, by the Bolzano-Weierstrass theorem, there exists at least one convergent subsequence. If there are finitely many convergent subsequences, then E is a finite set and thus, is closed.

The problem I am having is what if there are infinitely many convergent subsequences? The idea in my head is that the set E would contain elements which were isolated from one another. However, I am having a pretty hard time clarifying this idea in my head.

Any hints would be much appreciated! Thank you.

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# Real Analysis Question: Sequences and Closed Sets

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