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Let $X$ be connected metric space.

Let $U$ be an open subset of $X$.

Let $$Y=\big(\{0\}\times \overline U\big)\cup \big(\{1/n:n=1,2,3,...\}\times X\setminus U\big)$$ with subspace topology from $\mathbb R \times X$.

Is it necessarily true that $\{0\}\times \overline U\subseteq A$ whenever $A$ is a closed-and-open subset of $Y$ such that $A\cap (\{0\}\times \overline U)\neq\varnothing$?

What I do know is that the answer is YES if $\partial U$, the $X$-boundary of $U$, is compact. I have not been able to prove it otherwise, so maybe there is a counterexample?

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  • $\begingroup$ I think there is a proof as follows, but I have not worked out the details. Suppose there were a partition of $Y$ into disjoint clopen sets $A_1,A_2$ both containing elements of $\{0\}\times\bar U$. For both $i$, let $A'_i=\{x\in X \mid \forall\epsilon>0\;\exists\delta\in[0,\epsilon)\text{ s.t. }(\delta,x)\in A_i\}$. Then one can show (I think) that $A'_1$ and $A'_2$ are disjoint, open, nonempty, and cover $X$. $\endgroup$
    – MTyson
    Jan 31, 2018 at 4:05
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    $\begingroup$ The answer YES for the case of compact boundary implies the YES answer for a continuum-connected space. So, if a counterexample exists, it should not be continuum-connected. $\endgroup$
    – Taras Banakh
    Jan 31, 2018 at 6:55
  • $\begingroup$ @TarasBanakh Could you explain? The reason it holds for the compact boundary case is because given a $Y$-clopen set $A$, eventually, the "boundaries" of the halves (all homeomorphic to $\partial U$) must "match up" regarding their intersections with $A$ and its complement. And so $A$ will naturally correspond to a clopen subset of $X$. To MTyson: That is an interesting idea, but it's not immediately obvious to me that it works! $\endgroup$
    – D.S. Lipham
    Jan 31, 2018 at 17:31

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As Taras Banakh noted, the answer is yes if $X$ is a continuum-connected space.

I'll identify $X$ with $\{0\}\times X$ throughout. Suppose there were a partition of $Y$ into disjoint clopen sets $A_1,A_2$ both containing elements of $\overline U$. For both $i$ let $A'_i$ be the intersection of $X$ and the $\mathbb{R}\times X$-closure of $A_i$. It follows from the assumptions on the $A_i$ that the $A'_i$ are nonempty, closed, and cover $X$. It suffices to show that the $A'_i$ are disjoint.

Note that $A'_i\subseteq A_i\cup (\overline Y\setminus Y)=A_i\cup(X\setminus \overline U)$. In particular $A'_1\cap A'_2\cap\overline U=A_1\cap A_2\cap\overline U=\emptyset$.

Let $V$ be a connected component of $X\setminus U$. The $A_i$ are a clopen partition of $Y$, so each $\{1/n\}\times V$ is, as a connected component of $Y$, contained in exactly one of the $A_i$. As a result, each $A'_i$ must contain either all or none of $V$. Since $X$ is continuum-connected, $V$ is clopen and therefore there is a point $y$ in $\partial U\cap V$, which by the previous paragraph is contained in exactly one of the $A'_i$. Hence $A'_1\cap A'_2\cap V=\emptyset$, and moreover $A'_1\cap A'_2\cap (X\setminus U)=\emptyset$.

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    $\begingroup$ "Since $X$ is connected, there is a point $y$ in $\partial U\cap V$". This is false in general because there are strange connected sets like Cantor's Leaky Tent where the component of a point in an open set can be just the single point (there are even more horrible examples where the $\textit{quasi}$-components of certain open sets are single points). However, Your argument does work if $X$ is continuum-wise connected, so you proved Taras Banakh's comment above. 1 upvote for that! $\endgroup$
    – D.S. Lipham
    Jan 31, 2018 at 19:39
  • $\begingroup$ @David Thank you for the correction. I didn't know that. $\endgroup$
    – MTyson
    Jan 31, 2018 at 20:05

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