Closed set implies non-open set?

Question

Given $X$ a metric space and $E$ is a strict subset of $X$ that's non-empty and is closed in $X$, is it true that $E$ is not open?

My guess is no, considering I can form the metric space $X=(-1,1) \cup 2$ and then $E = (-1,1)$. Every limit point of $E$ in $X$ is a point of $E$ making $E$ closed in $X$, however, every point in $E$ is clearly an interior point as well, so $E$ is open in $X$.

Does anyone have anymore insight into this?

Thanks in advanced

If metric space $X$ is connected and $E$ is nonempty proper subset of $X$, then openness of $E$ implies $E$ is not closed.
If $X$ is topological space, existence of clopen nonempty proper subset of $X$ is equivalent to $X$ is not connected. — Hanul Jeon, May 10 '13 at 06:33

score 8 · Accepted Answer · answered May 10 '13 at 06:17

8

Sets which are both closed and open in a topological space are called clopen and the link gives several examples.

answered May 10 '13 at 06:17

Ragib Zaman

35,127

i totally racked my brain for 2 days thinking about this, glad i havent gone totally insane yet :P – DanZimm May 10 '13 at 06:23
4

@DanZimm It's only a matter of time. – Ragib Zaman May 10 '13 at 06:24
I'll be accepting as soon as SE lets me – DanZimm May 10 '13 at 06:26

score 4 · Answer 2 · edited Jun 12 '20 at 10:38

4

Depend on the metric space $X$.

Example 1 $X$ is $\mathbb R$ with usual topology. $E=[0,1]\subset X$, then $E$ is not open.

Example 2 $X$ is a discrete space. It always a metric space. Every set $E \subset X$ is open and closed.

edited Jun 12 '20 at 10:38

Community

1

answered May 10 '13 at 06:25

Paul

20,553

score 0 · Answer 3 · answered May 10 '13 at 13:56

0

Consider $\mathbb{R}\setminus{\{1,-1\}}$, $E:=(-1,1)$ is both open and closed in it

answered May 10 '13 at 13:56

Edoardo Lanari

3,564

Closed set implies non-open set?

3 Answers3