# Let $E \subset X$ be a connected set. Show that $\overline{E}$ is connected.

Let $$E \subset X$$ be a connected set. Show that $$\overline{E}$$ is connected.

Suppose that $$\overline{E}$$ is disconnected, then $$\overline{E} = A \cup B$$ where $$A \cap B = \emptyset$$ and $$A,B$$ are both open. Since $$E \subset \overline{E}$$ we must have that $$E \subset A$$ or $$E \subset B$$. Wlog assume that $$E \subset A$$, then $$E \cap B = \emptyset$$ i.e. $$E \subset B^c$$. This implies that $$\overline{E} \subset \overline{B^c} = B^c$$ but this happens if and only if $$\overline{E} \cap B = \emptyset$$? Is this a valid proof for this?

• Seems correct to me. Jul 26, 2022 at 16:59
• Your proof is fine. Just add one more line that $B=\emptyset$ Jul 26, 2022 at 17:21
• Jul 27, 2022 at 20:56

A topological space $$(X, \tau)$$ is connected iff no continuous map $$f:X\to\{0, 1\}$$ is onto. $$[$$ the target space is endowed by discrete topology $$]$$
Let $$f:\overline{E}\to \{0, 1\}$$ be a continuous function.
Goal: To show $$f$$ is not onto i.e $$f$$ is constant.
$$F:=f|_E : E\to \{0, 1\}$$ is also continuous and connectedness of $$E$$ implies $$F$$ is constant.
Since $$E\subset \overline{E}$$ is dense and the target space is Hausdorff , $$F=f$$ on $$E$$ implies $$F=f$$ on $$\overline{E}$$ implies $$f$$ is constant on $$\overline{E}$$.$$($$ see here and here $$)$$