(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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Real Analysis Question: Sequences and Closed Sets

**Physics Forums | Science Articles, Homework Help, Discussion**