Local connectedness in the boundary implies local connectedness in the closure 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!
 A: 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. 
