Exercise 2.20

We claim first that closures of connected sets are always connected. To show this consider any connected set $E$ and consider nonempty sets $A$ and $B$ where $Ē=A\cup B$. Now, let $Ĕ=Ē-E$, $C=A-Ĕ$, and $D=B-Ĕ$.

First we show that that $E=C\cup D$. So consider any $x\in E$. Then $x\in E\cup E^{\prime }=Ē$, from which it follows that $x\in A$ or $x\in B$ since $Ē=A\cup B$. Since $x\in E$ it follows that $x\notin Ĕ$. Then $x\in A-Ĕ=C$ or $x\in B-Ĕ=D$ so that $x\in C\cup D$. Thus $E\subseteq C\cup D$. Now consider any $x\in C\cup D$. If $x\in C$ then $x\in A$ and $x\notin Ĕ$. It follows that $x\notin Ē$ or $x\in E$, which is logically equivalent to $x\in Ē\to x\in E$. But since $x\in A$, $x\in Ē=A\cup B$ so that indeed $x\in E$. A similar argument shows that $x\in E$ if $x\in D$. Therefore $C\cup D\subseteq E$, hence $E=C\cup D$ as claimed.

Now since $E$ is connected and $E=C\cup D$, by the definition of connectedness $\bar{C}\cap D\ne \varnothing$ and $C\cap \bar{D}\ne \varnothing$. Thus there is an $x\in \bar{C}\cap D$ so that $x\in \bar{C}$ and $x\in D$. Since $x\in D$, $x\in B$ by definition, and since $x\in \bar{C}$, $x\in C\cup C^{\prime }$. If $x\in C$ then $x\in A$ by definition so that $x\in A\cup A^{\prime }=Ā$. On the other hand if $x\in C^{\prime }$ then $x$ is a limit point of $C$. In this case consider any neighborhood $N_r(x)$. Then there is a $y\in N_r(x)$ where $y\ne x$ and $y\in C$. So since $C=A-Ĕ$, $y\in A$. From this it follows that $x$ is also a limit point of $A$ since $N_r(x)$ was arbitrary and $y\ne x$. Thus $x\in A^{\prime }$ so that $x\in A\cup A^{\prime }=Ā$. Hence in either case $x\in Ā$ so since also $x\in B$, $x\in Ā\cap B$ so that $Ā\cap B\ne \varnothing$. A similar argument shows that $A\cap \bar{B}\ne \varnothing$, thereby showing that $Ē$ is connected.

However there is an example of a set in ${ℝ}^2$, that is connected but whose interior is not. Such an example can be constructed from the following simple sets:

Then $E=A\cup B\cup C$ is clearly connected and yet its interior is not since the interior of $C$ in ${ℝ}^2$ is the empty set, thereby separating the interiors of $A$ and $B$.