- #1
Szichedelic
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Homework Statement
Let {xn} be a sequence of real numbers. Let E denote the set of all numbers z that have the property that there exists a subsequence {xnk} convergent to z. Show that E is closed.
Homework 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.
The Attempt at a Solution
What I've done for this problem is I've split it up into several cases.
Case 1
{xn} 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
{xn} 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
{xn} 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.