Let $A$ be a closed subset of a connected top space $X$. If the $\text{Bd }A$ is connected, does it imply $A$ is also connected? I came across a question from Munkres:
Problem 24.11: If A is a connected subspace of X, does it follow that IntA and BdA are connected? Does the converse hold?
Thinking about the converse, I know that in general, if Bd A is connected, it doesn't necessarily mean that A is connected (e.g. the rationals). But what if X is connected, A is closed and the Bd A is connected? Would that then imply A is also connected?
 A: A discrete space with more than one point is a trivial counterexample. The boundary of any subset is empty, hence connected, yet there exist disconnected closed subsets of the space (e.g. any subset with more than one point).
If the space is connected however, we may prove that the result is true as follows. Suppose $C$ is a closed subset such that $\mathrm{Bd}\ C$ is connected and suppose $C$ is disconnected. Then there exist disjoint, nonempty open sets $U,V$ separating $C$. Being connected, the boundary of $C$ lies entirely within exactly one of $U$ and $V$, say it is $V$. Then it follows that $C-V=(\mathrm{Int}\ C)\cap U$. The left hand side is clearly closed while the right hand side is clearly open, thus there is a proper nonempty subset of $X$ that is both open and closed, contradicting connectedness.
A: Suppose that $A$ is closed in $X$, that $X$ and Bd$(A)$ are connected, but (toward a contradiction) that $A$ is disconnected.  So $A$ is the union of two disjoint, nonempty subsets $P$ and $Q$ that are closed in $A$ and therefore in $X$. Now consider any point $x$ in Bd$(A)$. Since $A$ is closed, it follows that $x\in A$ and therefore either $x\in P$ or $x\in Q$; without loss of generality, assume $x\in P$. Furthermore, $x$ is not in the interior of $A$ and therefore not in the interior of $P$.  So $x\in \text{Bd}(P)$. (Here and throughout this answer, boundaries and interiors are with respect to $X$.)  This shows that Bd$(A)$ is covered by Bd$(P)$ and Bd$(Q)$.  
I claim that Bd$(P)$ is actually included in Bd$(A)$; then the same will be true for Bd$(Q)$.  To verify the claim, consider any $y\in\text{Bd}(P)$.  It's in $P$ and therefore in $A$, so the only way it could avoid being in Bd$(A)$ would be if it's in the interior of $A$.  So suppose $y$ had an open neighborhood $U\subseteq A$. Since $y$ is not in the interior of $P$, there must be points in $U$ from $A-P=Q$.  The same goes for every neighborhood $V$ with $y\in V\subseteq U$, and therefore $y$ is in the closure of $Q$.  But $Q$ is closed and disjoint from $P$, so this is impossible for $y\in P$.  This completes the proof of the claim.
So now we have that Bd$(A)$ is the union of Bd$(P)$ and Bd$(Q)$.  These are disjoint (because $P\cap Q=\varnothing$), closed sets, yet Bd$(A)$ is connected.  So one of  Bd$(P)$ and Bd$(Q)$ has to be empty.  But that means one of $P$ and $Q$ is clopen in $X$, contrary to the assumption that $X$ is connected.
