(a) The closure of a disconnected set is disconnected?

(b) What about the interior of a disconnected set?

For (a) take $\mathbb{R}^{2}-\{x\text{-axis}\}$.

For (b) seems true. I just had an intuitive idea. If a set $S$ is disconnected, $S = A \cup B$ where $A\cap B = \emptyset$, $\overline{A}\cap B = \emptyset$ and $A \cap \overline{B} = \emptyset$. So, take the interior of $S$ is take the interior of $A$ and the interior of $B$, that is, $\text{int }S \subset\text{int }A \cup \text{int }B$ and satisfies the hypothesis for disconnectedness.

But I dont know if its correct. Can someone help me?

  • 1
    $\begingroup$ What if the interior was connected? $\endgroup$ – Randall Dec 22 '18 at 3:19
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    $\begingroup$ What about $(0,1)\cup\{2\}$? $\endgroup$ – Don Thousand Dec 22 '18 at 3:19
  • $\begingroup$ @DonThousand oh, its really simple! Thank you! $\endgroup$ – Corrêa Dec 22 '18 at 3:21
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    $\begingroup$ You can consider even a simpler disconnected set: $\{0,1\}$ $\endgroup$ – jjagmath Dec 22 '18 at 11:17

(a) is correct, but an even simpler example (in the reals) is $(0,1) \cup (1,2)$, which has closure $[0,2]$.

For (b) look no further than the reals again: the disconnected set $(0,1) \cup \{2\}$ has interior $(0,1)$, quite connected.


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