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

[Artusartos]

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

 

[pasmith]

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 $A_n = A_1$ for some $n > 1$.

The way to show that $b \in \bigcap A_n$ is to show that $b \in A_n$ for each $n \in \mathbb{N}$.

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

 

[Artusartos]

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

The way to show that $b \in \bigcap A_n$ is to show that $b \in A_n$ for each $n \in \mathbb{N}$.

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

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?

 

[pasmith]

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 $A_n$ which is a proper subset of $A_1$. If, by assumption, $b \notin A_n$, then it can only be because $A_n$ is such a proper subset, and there then exists an open neighbourhood $V$ of $b$ such that $V \cap A_n = \varnothing$.

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

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 $n \in \mathbb{N}$, the subsequence $\{a_n, a_{n+1}, \dots\}\subset A_n$ has limit $b$.

 

[jbunniii]

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 $A_n^c$?