# prove that if $E$ is connected and $E \subseteq F \subseteq \overline{E}$, then $F$ is connected.

Define a set $$A$$ to be disconnected iff there exist nonempty relatively open sets $$U$$ and $$W$$ in $$A$$ with $$U\cap W = \emptyset$$ and $$A = U\cup W.$$ Define a set $$A$$ to be connected iff it is not disconnected.(there are many equivalent definitions, but I want to prove this lemma using this one). Prove that if $$E$$ is connected and $$E\subseteq F \subseteq \overline{E},$$ then $$F$$ is connected.

Let $$U, W$$ be a separation for $$F$$. Find open sets $$O_U$$ and $$O_W$$ so that $$U = F \cap O_U$$ and $$W = F\cap O_W.$$ I claim that $$E\cap O_U, E\cap O_W$$ separate $$E$$. However, I'm unable to show that $$U' = E\cap O_U, W' =E\cap O_W \neq \emptyset$$ (I think this should be straightforward, but for some reason, I can't figure this out). Suppose $$U' = \emptyset.$$ Then $$E\cap O_U = \emptyset.$$ Since $$E \subseteq F = U\cup W = F\cap (O_U \cup O_W)\subseteq O_U\cup O_W,$$ we have that $$E \subseteq O_W,$$ so $$E\cap O_W = E\subseteq F\cap O_W = W\subseteq F\subseteq \overline{E}.$$ Observe that since $$E\cap O_U = \emptyset, F\cap O_U = (F\backslash E)\cap O_U\subseteq F\backslash E.$$

Similarly, $$W'\neq \emptyset.$$ Clearly, $$U', W'$$ are relatively open in $$E$$. Suppose $$U'\cap W' \neq \emptyset.$$ Let $$x\in U'\cap W'.$$ Then $$x\in E\cap O_U\cap O_W \subseteq F\cap O_U\cap O_W = U\cap W = \emptyset,$$ a contradiction. So $$U'\cap W' = \emptyset.$$ Also, $$U'\cup W' = (E\cap O_U)\cup (E\cap O_W) = E\cap (O_U \cup O_W)$$ and $$E\subseteq (O_U \cup O_W),$$ so $$U'\cup W' = E.$$

• If you have already proved that connected components are closed, you cam usit to prove your statement, and omit taking open sets. Jul 2, 2020 at 0:06

This follows from the following result:

Theorem: If $$Y$$ is a connected subset of a topological space $$X$$, then $$\overline{Y}$$ is connected.

Here is a short proof

Suppose $$\overline{Y}$$ is the union of two disjoint clopen sets $$A$$ and $$B$$ in $$\overline{Y}$$. Then $$A\cap Y$$ and $$B\cap Y$$ are clopen in $$Y$$. Hence, either $$A\cap Y=\emptyset$$ or $$B\cap Y=\emptyset$$. Suppose $$Y\cap B=\emptyset$$. Then $$Y\subset A$$ and so, $$\overline{Y}=\overline{A}=A$$ since $$A$$ is closed in $$\overline{Y}$$. Thus, $$B=\emptyset$$.

In your case, if $$E\subset F\subset \overline{E}$$ and $$E$$ is connected, then the closure of $$E$$ relative to $$F$$, given by $$\overline{E}\cap F=F$$, is connected.

Assume $$E \subseteq F \subseteq \overline{E} \subseteq X$$ and that we're working in the topology of $$E$$ relative to $$X$$. If $$F$$ is disconnnected, then $$\exists U, V \subseteq X$$ open (in $$X$$) such that $$F \subseteq U \cup V$$ and $$U \cap V \cap F = \emptyset$$.

But then $$U \cap V \cap E = \emptyset$$ and $$E \subseteq F \subseteq U \cup V$$, so $$U$$ and $$V$$ are a separation for $$E$$, which is connected. Thus, without loss of generality, $$E \subseteq U$$ and $$E \cap V = \emptyset$$.

But then $$E \subseteq X \setminus V$$, which is closed (because $$V$$ is open), and since the closure of $$E$$ is the intersection of all closed sets containing $$E$$, that means $$\overline{E} \subseteq X \setminus V$$, so $$\overline{E} \cap V = \emptyset$$ and since $$F \subseteq \overline{E}$$, we also have $$F \cap V = \emptyset$$ so that $$U, V$$ do not separate $$F$$.