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Suppose $A\subset \mathbb{R}^2$ is closed and totally disconnected, and suppose $a,b \in A$. Is it possible to find a compact subset $C$ of the plane, which is disjoint from $A$, and which separates $a$ and $b$? (That is, $a$ and $b$ are in different components of $\mathbb{R}^2-C$.)

If $A$ has no points other than $a$ and $b$, we could let $C$ be a circle around $a$ so that $b$ is on the outside. But if $A$ has additional points, the circle could intersect them, so a different choice has to be made.

We can assume for this problem that if $U$ is any connected open set in $\mathbb{R}^2$, then $U-A$ is also connected. If that helps.

This problem arose while I was trying to prove this one I submitted earlier.

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This is trickier than I thought. :D –  BBischof May 3 '11 at 2:23
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up vote 1 down vote accepted

Ok, here's a shot at it.

Note that it's sufficient to find a separation of $A$ of the form $A \subset U \cup V$ where $U,V$ are open, $U \cap V = \emptyset$, $a \in U$, $b \in V$, and $U$ is precompact. If so, then $C = \partial U$ will satisfy our purpose. (Verify!) The precompactness of $U$ is the only hard part here.

Let $W$ be a precompact neighborhood of $a$ with $b \notin \bar{W}$. For each $x \in A \cap \partial W$, by total disconnectedness we can write $A \subset U_x \cup V_x$ where $U_x, V_x$ are open and disjoint, $a \in U_x$, and $x \in V_x$. The $V_x$ are an open cover of $A \cap \partial W$, so we can find an finite subcover $V_{x_i}$. Then $U = \bigcap_i U_{x_i}$, $V = \bigcup_i V_{x_i}$ are open and disjoint, $A \subset U \cup V$, $a \in U$, and $A \cap \partial W \subset V$. So if we set $U' = U \cap W$ and $V' = V \cup \bar{W}^c$, we have the desired separation.

This should still work if $\mathbb{R}^2$ is replaced by any locally compact Hausdorff space.

Any bugs?

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This is a very nice solution, thank you very much. –  alephzero314 May 3 '11 at 7:36
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