# The interior of a disconnected set is disconnected?

(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?

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