TITLE: Is it true that if $X$ is connected, then for every nonempty proper subset $A$ of $X$, we have $\mathbf{Bd} \ne \emptyset$
QUESTION [6 upvotes]: Is it true that if $X$ is connected, then for every nonempty proper subset $A$ of $X$, we have $\mathbf{Bd} \ne \emptyset$? Does the converse hold?
I start by trying to understand what $\mathbf{Bd}$ means in "formal terms" (i.e. $\bar{A} \cap \overline{X-A}$ is the 'boundary' of $A \in X$)?
This means I'm supposed to figure out if the boundary of subset of a connected space is empty or not, right? I think that if a set is connected then it contains no partition into two sets, so there would exist points that are both in $A$ and in the complement of $A$ in $X$. This would be an informal definition of a boundary. (A boundary contains points in a set $A$ and points outside of it, right?) 
What I mean is, if $X$ is connected, and $A \subset X$ is connected, then $A$ could have a boundary which means that technically, $X$ cannot be partitioned into $A$ and $A$-complement without making the boundary of $A$ an empty set.
Right?

REPLY [4 votes]: Be careful: any set with more than one element can be partitioned into two non-empty sets; a connected set cannot be partitioned into two non-empty relatively open sets. And in your last paragraph it’s the closures of $A$ and $X\setminus A$ whose disjointness makes the boundary of $A$ empty.
The easiest way to prove the result is to prove the contrapositive. Suppose that $A$ is a non-empty proper subset of a space $X$, and suppose further that $\operatorname{bdry}A=\varnothing$. As you say, $\operatorname{bdry}A=(\operatorname{cl}A)\cap\operatorname{cl}(X\setminus A)$, so $\operatorname{cl}A$ and $\operatorname{cl}(X\setminus A)$ are disjoint closed sets whose union is $X$. Each of them is therefore the complement of a closed set, so each is open. Moreover, $\operatorname{cl}A\supseteq A\ne\varnothing$ and $\operatorname{cl}(X\setminus A)\supseteq X\setminus A\ne\varnothing$, so $\{\operatorname{cl}A,\operatorname{cl}(X\setminus A)\}$ is a partition of $X$ into non-empty open sets, and $X$ therefore is not connected.

REPLY [3 votes]: Your intuition about their equivalence is spot on. The following conditions can be shown to be equivalent:

$X$ is connected.
If $A$ is a subset of $X$ that is both closed and open, then $A=\emptyset$ or $A=X$.

Observing that a subset of $X$ has empty boundary if and only if it is both closed and open, the equivalence follows immediately.
It's worth noting that any connected space with more than one point can be partitioned into two disjoint non-empty sets. However, it cannot be partitioned in this way with both sets open (or both sets closed, which is the same thing).