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Given a locally connected topological space $X$, and $A \subset X$ such that the boundary of $A$ is locally connected, is true that the closure of $A $ is also locally connected? I tried hard to prove this without success. Any help? Thank you!

Forever Mozart
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Andrés Ibáñez Núñez
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    What do you mean by locally connected? – André Porto May 13 '17 at 12:15
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    A topological space is locally connected if it has a basis formed by connected open sets, that is, if for every open set U and $x \in U$, there exists a connected open set V such that $x \in V \subset U$. Above boundary of A and clausure of A are considered with the subspace topology. – Andrés Ibáñez Núñez May 13 '17 at 12:18
  • What about the topological sine curve? – Henno Brandsma May 13 '17 at 14:53
  • @HennoBrandsma As far as I know, the topologist's sine curve is its own boundary and it is not locally connected, so it would not work as a counterexample. Am I correct? – Andrés Ibáñez Núñez May 13 '17 at 16:04
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    @AndrésIbáñezNúñez take the full topologist's sine as a space in its own right. The right hand part $A$ has ${0}\times [-1,-1]$ as its boundary, which is locally connected, but $\overline{A}$ is not. – Henno Brandsma May 13 '17 at 16:34
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    @HennoBrandsma. Letting X be the topologist's sine curve does not fit the hypothesis of the Q, that X be locally connected. – DanielWainfleet May 13 '17 at 19:17

1 Answers1

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It is true.

Under your assumptions we show that $\overline A$ is locally connected.

First note that $\overline A\setminus \partial A$ is open in $X$, so it follows from local connectedness of $X$ that $\overline A$ is locally connected at each point of $\overline A\setminus \partial A$.

Now let $x\in \partial A$ and let $U'$ be any $\overline A$-open neighborhood of $x$. We want to find a connected $\overline A$-open neighborhood of $x$ that is contained in $U'$.

Since $\partial A$ is locally connected, there is a connected $\partial A$-open $V'\subseteq U'\cap \partial A$ with $x\in V'$.

Let $V$ be an open subset of $X$ such that $V\cap \partial A=V'$.

For each $y\in V'$ let $W_y$ be a connected open subset of $X$ such that $y\in W_y\subseteq U\cap V$, where $U$ is open in $X$ such that $U\cap \overline A=U'$.

Let $W=\bigcup _{y\in V'}W_y$. It is easy to see that $W$ is connected.

$W':=W\cap \overline A$ is our desired set. Clearly $W'$ is open in $\overline A$ and $W'\subseteq U'$. To complete the proof we just need...

Claim: $W'$ is connected.

Note that $W'=V'\cup (W'\setminus V')$. If $C$ is a relatively clopen subset of $W'$ that meets $V'$, then $V'\subseteq C$. If $C$ does not also contain $W'\setminus V'$, then the two sets $$W'\setminus C$$ $$C\cup (W\setminus \overline A)$$ form a disconnection of $W$, a contradiction. Thus $W'\subseteq C$. This proves the claim.

Andrés Ibáñez Núñez
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Forever Mozart
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