If $A$ is a connected subspace, then any subspace containing $A$ that sits inside Cl($A$) is also connected.

Question

Let $A_{\tau_A}$ be a connected subspace of a space $X_{\tau}$, and let $B$ be a subset of $X$ with $A\subseteq B\subseteq \text{Cl($A$)}$. Then $B$ is also connected in the subspace topology.

Proof: First, if $B = \: \text{Cl}(A)$, then by Theorem 6.2.3, $B$ is connected in the subspace topology, and if $B = A$, then $B$ is connected in the subspace topology by assumption. Hence, suppose that $A \subset B \subset \: \text{Cl}(A)$. By way of contradiction, suppose that $B$ is disconnected in the subspace topology. Then $B$ can be written as a disjoint union of sets $C, D$ both proper, nonempty subsets of $B$, where $C = U \cap B, D = V \cap B$, where $U, V \in \tau$...

At this point, I am getting stuck and not sure how to proceed. Thanks for any help.

Answer 1

Since $B \subseteq Cl(A)$, $U \cap B \neq \emptyset \implies U \cap A \neq \emptyset$. Similarly for $V$. Note we use the openness of $U,V$ essentially here. Thus if $U,V$ are a disconnection on $B$, they are a disconnection on $A$, a contradiction.

Answer 2

You don’t have to work that hard: $\operatorname{cl}_BA=B$, so you can apply your Theorem 6.2.3 to $A$ in the space $B$ to conclude that $B$ is connected.