# limit points and closure for connected sets

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

###### Theorem 1.

Suppose $A$ is a connected set in a topological space. If $A\subseteq B\subseteq\overline{A}$, 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 dense subspace in $X$ which is connected.

Two touching closed balls in $\mathbbmss{R}^{2}$ shows that this theorem does not hold for the interior. Along the same lines, taking the closure does not preserve separatedness.

###### Proof.

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

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

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

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

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 $\emptyset\neq\tilde{U}\cap\tilde{V}=(R\cap S)\cap A$. Then, by the properties of the closure operator,

 $\emptyset\neq\overline{(R\cap S)\cap A}\supseteq(R\cap S)\cap\overline{A}% \supseteq(R\cap S)\cap B=\emptyset.$

Title limit points and closure for connected sets LimitPointsAndClosureForConnectedSets 2013-03-22 15:17:56 2013-03-22 15:17:56 matte (1858) matte (1858) 7 matte (1858) Theorem msc 54D05