$A$ compact $\implies$ $\overline{A}$ compact?

Question

Let $(X,\tau)$ a topological space

If $A \subset X$ is a compact set, does this imply that $\overline{A}$ is a compact set?

score 4 · Answer 1 · edited Nov 06 '18 at 23:13

4

No. For example, let $X=\mathbb{R}$ with topology $$ \tau=\{A\cup\{0\}\colon A\text{ is open in usual topology}\}\cup\{\varnothing\}. $$ The singleton $\{0\}$ is obviously compact, but $\overline{\{0\}}=\mathbb{R}$ is not.

edited Nov 06 '18 at 23:13

irchans

1,915

answered Nov 06 '18 at 22:59

user10354138

33,239

$(1,2)$ is open in your topology and doesn't contain $0$ so $1,5 \notin \overline{{0}}$ – Maxime Ramzi Nov 06 '18 at 23:00
oops, corrected – user10354138 Nov 06 '18 at 23:02

score 4 · Accepted Answer · answered Nov 06 '18 at 22:59

The answer is yes if $X$ is Hausdorff (because then $A$ is closed), but no in general.

Consider $X=\mathbb{N}$ with the following topology : its opens are the empty set and the sets that contain $0$. Then $X$ is not compact because $\displaystyle\bigcup_{n\in \mathbb{N}}\{0,n\}$ is an open covering with no finite subcover; $\{0\}$ is compact; and the closure of $\{0\}$ is $X$, so $\overline{\{0\}}$ is not compact.

$A$ compact $\implies$ $\overline{A}$ compact?

2 Answers2