1

Having trouble on proving this, I was thinking of doing the contrapositive. Suppose $\overline{A}$ is connected i.e. $(\forall B, C \subseteq \overline{A}) (B \cup C \neq \overline{A})$ or $(\overline{B} \cap C \neq \emptyset)$ or $(B \cap \overline{C} \neq \emptyset)$. I have written much more of what I tried on paper, but it seems pointless to put what I tried. I pretty much assumed to show that $A$ is connected, let $A = B_1 \cup C_1$ where $B_1$ and $C_1$ are both subsets for $B$ and $C$ from above.

3j iwiojr3
  • 728
  • 5
  • 11
  • 1
    From How to ask a good question: "Your question should be clear without the title. After the title has drawn someone's attention to the question by giving a good description, its purpose is done. The title is not the first sentence of your question, so make sure that the question body does not rely on specific information in the title." – jjagmath Jul 17 '22 at 23:32

4 Answers4

4

This is wrong. $\mathbb Q$ is disconnected but its closure is connected.

Kavi Rama Murthy
  • 359,332
3

$\Bbb R \setminus \{0\}$ is a very simple counterexample

jjagmath
  • 22,582
1

As the other answers have already pointed out, this claim is wrong. But what is true is, that for a topological space $X$ and a connected subspace $A\subset X$, any subspace $B\subset X$ with $A\subseteq B\subseteq\overline{A}$ is connected. For a proof see here, here or here.

Samuel Adrian Antz
  • 2,214
1

$A=(0, 1) \cup (1, 2) $ is disconnected subset of $\Bbb{R}$ with euclidean topology. But $\overline{A}=[0, 1]\cup[1, 2]=[0, 2]$ is connected.

SoG
  • 13,327