limit points and closure for connected sets

The below theorem shows that adding limit points to a connectedset preserves connectedness.

Theorem 1.

Suppose $A$ is a connected set in a topological space.If $A\mathrm{\subseteq }B\mathrm{\subseteq }\overline{A}$, then $B$ is connected.In particular, $\overline{A}$ is connected.

Thus, one way to prove that a space $X$ is connected is to find a densesubspace in $X$ which is connected.

Two touching closed balls in ${ℝ}^2$ shows that this theorem does not holdfor the interior. Along the same lines, taking the closure does notpreserve separatedness.

Proof.

Let $X$ be the ambient topological space.By assumption, if $U,V\subseteq A$ are open and $U\cup V=A$, then$U\cap V\ne \mathrm{\varnothing }$.To prove that $B$ is connected, let $U,V$ be open sets in$B$ such that $U\cup V=B$ and for acontradition, suppose that $U\cap V=\mathrm{\varnothing }$.Then there are open sets $R,S\subseteq X$ such that

$$U=R\cap B,V=S\cap B.$$

It follows that $(R\cup S)\cap B=B$and $(R\cap S)\cap B=\mathrm{\varnothing }$.Next, let $\tilde{U},\tilde{V}$ be open sets in $A$ defined as

$$\tilde{U}=R\cap A,\tilde{V}=S\cap A.$$

Now

$$A=B\cap A\subseteq (R\cup S)\cap A\subseteq A$$

and as $(R\cup S)\cap A=\tilde{U}\cup \tilde{V}$, it follows that$\mathrm{\varnothing }\ne \tilde{U}\cap \tilde{V}=(R\cap S)\cap A$.Then, by the properties of the closure operator,

$$\mathrm{\varnothing }\ne \overline{(R\cap S)\cap A}\supseteq (R\cap S)\cap \overline{A}\supseteq (R\cap S)\cap B=\mathrm{\varnothing }.$$

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