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The following questions showed up on a previous exam of mine, but I didn't know how to do it at the time. I'll post my work after the statement of the question, but was wondering if anyone could help me out either with a different proof or with completing my proof.

Let $X$ be a path-connected space. Given a subset $A \subset X$ and two points $a,b \not \in A$, $A$ separates $a$ from $b$ if $a$ and $b$ lie in different path components of $X \setminus A$. Let $A_1 \supset A_2 \supset \ ...$ be a decreasing sequence of closed subset of $X$ that separate two fixed points $a$ and $b$. Prove that $A = \bigcap_i A_i$ separates $a$ from $b$.

I think I can visualize this in a particular case: In $\mathbb{R}$, we can take the nested, closed intervals $[-1,1] \supset [-\frac{1}{2}, \frac{1}{2}] \supset \ ...$ that separate, say, $-2$ from $2$. By the nested interval theorem, the intersection is non-empty (i.e. $\{0\}$, which separates $-2$ and $2$). I don't know how to generalize this idea. How I thought to approach this: Let $\gamma$ be an arbitrary path from $a$ to $b$, and consider the image set $\gamma[0,1]$. I'd like to show that the intersection of $\gamma[0,1] \cap A$ is non-empty. Not sure where I'd go from here, but it was the first thing to come to mind. Any help is appreciated!

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In the first line of the shaded box, do you mean $a,b\notin A$ ? –  Stefan Hamcke Jan 15 at 19:51
3  
The trace of a path is a compact set. Does that ring a bell? –  Daniel Fischer Jan 15 at 19:55
    
You should be able to get there using nets, too. –  A.P. Jan 15 at 20:01
    
Thanks, Stefan. Corrected! –  user121410 Jan 15 at 20:16

1 Answer 1

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Assume that there is a path from $a$ to $b$, call it $p$, which avoids $A$. As each $A_i$ seperated $a$ and $b$, the path $p$ intersects $A_i$. Now $B_i:=A_i\cap p(I)$ is a decreasing sequence of closed subsets of $p(I)$ having the finite intersection property (since the intersection of a finite number of $B_i$ is just the smallest of these). By compactness of $p(I)$ it follows that $\bigcap_i p(I)\cap A_i=p(I)\cap A$ is non-empty. So there can be no path in $X\setminus A$ from $a$ to $b$.

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Thanks for posting! I think I understand this argument, save for two points: How do I show that $p(I)$ is 1) closed and 2) compact? I'm assuming I need 1) to show that the B_i's are closed. –  user121410 Jan 15 at 20:15
    
@user121410: 1) You do not need the closedness of $B_i$ in $X$, only the closedness in the subspace topology of $p(I)$ is important. 2) $p(I)$ is the image of the compact space $I$ under a continuous map. –  Stefan Hamcke Jan 15 at 20:18
    
Oh yes! Thank you! Not sure why I forgot the compactness part. Thanks, Stefan! –  user121410 Jan 15 at 20:32

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