I am trying to prove that if $X$ is a connected subset of a metric space $M$, then $\bar{X}$ is connected as well, that is, the closure of $X$ is connected.
My proof:
I will do a proof by contradiction. Suppose that $\bar{X}$ is disconnected. Then, there exist open set $U,V \subset \bar{X}$ which are relatively open in $X$, non empty and disjoint. Now, $U' = U \cap X$ and $V' = V \cap X$ are relatively open in $X$. Now, I was told by my book that to obtain the contradiction, we would like to show that $U', V'$ are both non-empty. I do not understand why showing $U', V'$ are both non-empty will lead to a contradiction. Anyone know? Thank you!
Best Answer
So suppose $X$ is connected and $\overline{X}$ is disconnected, so there are relatively open subsets $U,V$ of $\overline{X}$ that are both non-empty, disjoint and $U \cup V = \overline{X}$. So there are open sets $U',V'$ in $M$ such that $U = U' \cap \overline{X}$ and $V = V' \cap \overline{X}$, by the definition of relative (or subspace) topology.
Claim: $U' \cap X \neq \emptyset$. This follows as otherwise every point of $U'$ is not in $\overline{X}$, but we know that all points of $\emptyset \neq U \subset U'$ are in $\overline{X}$... Symmetrically, $V' \cap X \neq \emptyset$ as well.
But then $U' \cap X$ and $V' \cap X$ are disjoint, non-empty (see above) and cover $X$ (as $X = U \cup V \subset U' \cup V'$). This contradicts that $X$ is connected.