Real Analysis Question: Sequences and Closed Sets

In summary, the problem proves that the set E, defined as the set of all numbers z that have the property of having a subsequence {xnk} convergent to z, is closed. This is shown by splitting the problem into three cases: when {xn} is convergent, unbounded and divergent, and bounded and divergent. In each case, E is either a finite set or the empty set, which are both closed. The problem also considers the possibility of infinitely many convergent subsequences, but it is not necessary to prove the closure of E. Instead, it is enough to show that the complement of E, which is the set of all numbers that do not have a convergent subsequence, is
  • #1
Szichedelic
29
0

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.
 
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  • #2
Try showing that the complement of E is open. You don't really need all those cases.
 
  • #3
So the complement of E would be all such numbers that don't have a subsequence which converges to it?
 
  • #4
Szichedelic said:
So the complement of E would be all such numbers that don't have a subsequence which converges to it?

Sure. If you can show that set is open, then E is closed.
 

1. What is a sequence in real analysis?

A sequence in real analysis is a list of numbers that are indexed by the natural numbers (1, 2, 3, ...). Each number in the sequence is called a term, and the index indicates the position of the term in the sequence. For example, the sequence {1, 4, 9, 16, ...} is a sequence of perfect squares with terms in the first, second, third, and fourth positions.

2. How is a sequence represented mathematically?

A sequence can be represented mathematically as a function from the natural numbers to the real numbers. In other words, the sequence {1, 4, 9, 16, ...} can be represented as the function f(n) = n^2, where n is the index of the term in the sequence.

3. What is the limit of a sequence?

The limit of a sequence is the value that the terms in the sequence approach as the index of the terms increases without bound. In other words, as the index approaches infinity, the terms get closer and closer to the limit value. The limit of a sequence can be finite or infinite.

4. How do you determine if a sequence is convergent or divergent?

A sequence is convergent if its limit exists and is a finite value. In other words, the terms in the sequence approach a single value as the index increases. A sequence is divergent if its limit does not exist or is infinite, meaning the terms in the sequence do not approach a single value as the index increases.

5. What is a closed set in real analysis?

A closed set in real analysis is a set that contains all of its limit points. In other words, every sequence of points in the set that converges to a limit point is also contained in the set. This can also be defined as a set whose complement is open.

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