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Suppose $X$ is a topological space. Now let $A_1 \supset A_2 \supset \cdots$ be a sequence of closed subsets of $X$. Suppose $a_i \in A_i$ $\forall i$ and $a_i \rightarrow b$. Prove that $b \in \bigcap A_i$. Now I'm not so sure if I should do this, but do I need to use the finite intersection property somehow to get to the conclusion? If so, how exactly? I would appreciate all the help I can get, thank you.

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@Asaf: It’s a sequence whose $i$-th term is in $A_i$. –  Brian M. Scott Jan 12 '13 at 17:19
    
@Brian: I forget that the sets are decreasing... –  Asaf Karagila Jan 12 '13 at 17:22
    
Hint: $\cap A_i$ is closed. What if $b\notin\cap A_i$? –  David Mitra Jan 12 '13 at 17:22

2 Answers 2

up vote 2 down vote accepted

This is less by finite intersection properties, and more by the definition of closed sets and convergence.

Recall that $a_i\to b$ if for every open set $U$ such that $b\in U$ there is some $i_0$ such that for all $i>i_0$ we have $a_i\in U$. This means that by removing a [proper] initial segment of the sequence we do not affect the convergence of it.

Now recall that $b\in\operatorname{cl}(A)$ if and only if every open set $U$ which contains $b$ has a nonempty intersection with $A$. If $A$ is already closed then $\operatorname{cl}(A)=A$.

What you need to write, if so, is that whenever $U$ is an open set such that $b\in U$ then $B\cap A_i\neq\varnothing$ for all $i$, and therefore $b\in A_i$ for all $i$, and so $b\in\bigcap A_i$.

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Fix $n\in\Bbb N$; then $a_k\in A_n$ for every $k\ge n$, $\langle a_k:k\ge n\rangle\to b$, and $A_n$ is closed, so $b\in A_n$. But $n\in\Bbb N$ was arbitrary, so $b\in A_n$ for every $n\in\Bbb N$, and the result is immediate.

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