The Connectedness of the Closure of a Set

# The Connectedness of the Closure of a Set

Recall from the Connected and Disconnected Sets in Topological Spaces page that if $X$ is a topological space and $A \subseteq X$ then $A$ is said to be connected if the topological subspace $A$ is connected, and similarly, $A$ is said to be disconnected if the topological subspace $A$ is disconnected.

We will now look at a nice theorem which says that if $A$ is any connected set in a topological space then $\bar{A}$ is also a connected set. Of course, this is trivially true when $A$ itself is a closed set in $X$ (since then $A = \bar{A}$) but of course the result is more interesting when $A$ is not closed since then a connected $A$ is properly contained in a larger connected set $\bar{A}$.

 Theorem 1: Let $X$ be a topological space and let $A \subseteq X$. If the subspace $A$ is connected then the closure $\bar{A}$ is also connected.
• Proof: Let $A$ be a connected topological subspace and assume that $\bar{A}$ is disconnected. We will show that this leads to a contradiction.
• If $\bar{A}$ is disconnected then there exists open sets $B, C \subset \bar{A}$ where $B, C \neq \emptyset$, $B \cap C = \emptyset$ and $\bar{A} = B \cup C$.
• Since $\bar{A} = B \cup C$, if we take the closure of both sides of this equality we have that:
(1)
\begin{align} \quad \bar{\bar{A}} = \bar{A} = \overline{B \cup C} \overset{*} = \bar{B} \cup \bar{C} \end{align}
• Now since $B$ and $C$ are clopen sets in $\bar{A}$ we have that in fact $B = \bar{B}$ and $C = \bar{B}$ (since $B$ and $C$ are closed sets in $\bar{A}$).
• Since $A \subseteq \bar{A}$, write $A = (A \cap B) \cup (A \cap C)$. Then $A \cap B$ and $A \cap C$ are both open in $A$. Since $A$ is connected we must have that $(A \cap B) = \emptyset$ or $(A \cap C) = \emptyset$ (otheriwse then $\{ A \cap B, A \cap C \}$ would be a separation of $A$.
• Suppose that $A \cap B = \emptyset$. Then $A = A \cap C$. This implies that $A \subseteq C$, i.e., $A \subseteq \bar{C}$. Taking the closure of both sides shows that $\bar{A} \subseteq \bar{C}$. But then $\bar{A} = \bar{C}$, a contradiction.
• Similarly, if we instead suppose that $A \cap C = \emptyset$ then $A = A \cap B$. This implies that $A \subseteq B$, i.e., $A \subseteq \bar{B}$. Taking the closure of both sides shows that $\bar{A} \subseteq \bar{B}$. But then $\bar{A} = \bar{C}$, once again, a contradiction.
• Therefore the assumption that $\bar{A}$ was disconnected is false. Hence if $A$ is a connected set then the closure $\bar{A}$ is also connected. $\blacksquare$