Nested sequence of closed sets and convergence in a topological space.

In summary: Can ##a_i## be in ##A_n##?In summary, the proof first assumes that ##b \not\in \bigcap A_i## and then shows that this leads to a contradiction, implying that ##b \in \bigcap A_i##. This contradiction is reached by showing that if ##b \not\in \bigcap A_i##, then there exists an open neighborhood ##V## of ##b## such that ##V \cap A_n = \emptyset## for some ##n \in \Bbb{N}##. However, this is not possible because ##a_i## converges to ##b##, meaning that ##V## must contain all but finitely many elements of
  • #1
Artusartos
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Homework Statement



Let ##X## be a topological space. Let ##A_1 \supseteq A_2 \supseteq A_3...## be a sequence of closed subsets of ##X##. Suppose that ##a_i \in Ai## for all ##i## and that ##a_i \rightarrow b##. Prove that ##b \in \cap A_i##.

Homework Equations


The Attempt at a Solution



Suppose, for contradiction, that ##b \not\in \cap A_i##. Then there exists ##n \in \Bbb{N}## with ##b \not\in A_n##. But then ##b \not\in A_i## for all ##i \geq n##.

Since ##A_n## is closed, we can pick an open neighborhood ##V## of ##b \in A_1## with ##V \cap A_n = \emptyset##. By the definition of convergence, ##V## needs to contain all ##a_i## with ##i \geq N## for some ##N##. This is not possible because only a finite number of the ##a_i## can be contained ##V##, since ##b \not\in A_i## for all ##i \geq n##.

Is my answer correct?

Thanks in advance
 
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  • #2
Artusartos said:

Homework Statement



Let ##X## be a topological space. Let ##A_1 \supseteq A_2 \supseteq A_3...## be a sequence of closed subsets of ##X##. Suppose that ##a_i \in Ai## for all ##i## and that ##a_i \rightarrow b##. Prove that ##b \in \cap A_i##.



Homework Equations





The Attempt at a Solution



Suppose, for contradiction, that ##b \not\in \cap A_i##. Then there exists ##n \in \Bbb{N}## with ##b \not\in A_n##. But then ##b \not\in A_i## for all ##i \geq n##.

Since ##A_n## is closed, we can pick an open neighborhood ##V## of ##b \in A_1## with ##V \cap A_n = \emptyset##.

This is false. The hypotheses allow successive subsets to be equal, so it may be that [itex]A_n = A_1[/itex] for some [itex]n > 1[/itex].

The way to show that [itex]b \in \bigcap A_n[/itex] is to show that [itex]b \in A_n[/itex] for each [itex]n \in \mathbb{N}[/itex].

You know that [itex]A_n[/itex] is closed, so it contains all its limit points. Does there exist a sequence lying within [itex]A_n[/itex] whose limit is [itex]b[/itex]?
 
  • #3
pasmith said:
This is false. The hypotheses allow successive subsets to be equal, so it may be that [itex]A_n = A_1[/itex] for some [itex]n > 1[/itex].

The way to show that [itex]b \in \bigcap A_n[/itex] is to show that [itex]b \in A_n[/itex] for each [itex]n \in \mathbb{N}[/itex].

You know that [itex]A_n[/itex] is closed, so it contains all its limit points. Does there exist a sequence lying within [itex]A_n[/itex] whose limit is [itex]b[/itex]?

If ##A_n = A_1## for any ##n>1##, then we automatically get a contradiction, because ##a_i \rightarrow b## implies that ##b## is in the closure of ##A_1##. However, ##\bar{A_1} = A_1## and ##b \not\in A_1##. A contradiction, right?
 
  • #4
Artusartos said:
If ##A_n = A_1## for any ##n>1##, then we automatically get a contradiction, because ##a_i \rightarrow b## implies that ##b## is in the closure of ##A_1##. However, ##\bar{A_1} = A_1## and ##b \not\in A_1##. A contradiction, right?

In order for your proof to work, you need there to exist some [itex]A_n[/itex] which is a proper subset of [itex]A_1[/itex]. If, by assumption, [itex]b \notin A_n[/itex], then it can only be because [itex]A_n[/itex] is such a proper subset, and there then exists an open neighbourhood [itex]V[/itex] of [itex]b[/itex] such that [itex]V \cap A_n = \varnothing[/itex].

The only circumstance in which you can't do this is when [itex]A_n = A_1[/itex] for all [itex]n \in \mathbb{N}[/itex], in which case trivially [itex]b \in A_1 = \cap A_n[/itex].

Your proof just needs to be explicit about these points.

My other point is that there is a simpler direct proof which doesn't have this difficulty: for each [itex]n \in \mathbb{N}[/itex], the subsequence [itex]\{a_n, a_{n+1}, \dots\}\subset A_n[/itex] has limit [itex]b[/itex].
 
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  • #5
Artusartos said:
Since ##A_n## is closed, we can pick an open neighborhood ##V## of ##b \in A_1## with ##V \cap A_n = \emptyset##.
Two comments:
1. What do you mean by "an open neighborhood ##V## of ##b \in A_1##"? Are you claiming that ##b \in A_1##? On what basis? And why does your proof need this to be true anyway?
2. If ##b \not\in A_n##, then ##b \in A_n^c##. Since ##A_n## is closed, ##A_n^c## is an open neighborhood of ##b##. You don't need to introduce a new set ##V##. If ##i \geq n##, can ##a_i## be in ##A_n^c##?
 
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1. What is a nested sequence of closed sets?

A nested sequence of closed sets is a sequence of sets in which each set is a subset of the next one, and all sets in the sequence are closed. This means that the limit of the sequence is contained within each set, and the sequence continues to "nest" or become smaller as it progresses.

2. How are nested sequences of closed sets used in topology?

Nested sequences of closed sets are useful in topology because they allow for the definition of convergence in a topological space. In other words, a sequence of points in a topological space converges if and only if it is contained in a nested sequence of closed sets.

3. What is the limit of a nested sequence of closed sets?

The limit of a nested sequence of closed sets is the intersection of all sets in the sequence. This is because the sets in the sequence become smaller and smaller, and the limit is the point or points that are contained in all of the sets in the sequence.

4. How does the concept of convergence relate to nested sequences of closed sets?

The concept of convergence in a topological space is closely related to nested sequences of closed sets. A sequence of points in a topological space converges if and only if it is contained in a nested sequence of closed sets, and the limit of the sequence is the intersection of all sets in the sequence.

5. Can nested sequences of closed sets be used to prove continuity in topology?

Yes, nested sequences of closed sets can be used to prove continuity in topology. In fact, one of the equivalent definitions of continuity in topology is that the pre-image of every nested sequence of closed sets is a nested sequence of closed sets. This means that if a function is continuous, then the inverse image of a sequence of closed sets will also be a sequence of closed sets, allowing for the use of nested sequences to prove continuity.

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