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?

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

1 Answer 1



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 $) $


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