0
$\begingroup$

It is clear how to proof that the set $A$ is open. I just need to find some sequence which elements belong to $A$ while its limit does not.

It is also clear that in order to show that the set $A$ is closed I need to verify that its complement $\overline{A}$ is open.

But sometimes while the set $A$ complement $\overline{A}$ is open the complement $\overline{\overline{A}}$ of this complement $\overline{A}$ is also open. So we get clopen set.

I want to show that the set $A$ is closed but not clopen. Is it enough to show that $\overline{A}$ is open and $\overline{\overline{A}}$ is closed? And why it is not follows immidiately that $\overline{\overline{A}}$ is closed if I had previously shown that $A$ is closed?

Maybe it exist some very simple while sophisticated example of clopen set? It is not clear to me how the set $\overline{\overline{A}}$ could be open while the set $A$ is closed while them should be the same sets.

Will be very greatfull for help!

$\endgroup$
  • 1
    $\begingroup$ I'm confused with your notation.Do you mean $\overline{A} = X \setminus A$ ? where $X$ is a whole space $\endgroup$ – Chinnapparaj R Sep 22 '18 at 9:40
  • $\begingroup$ Yes, that is right $\endgroup$ – Bogdan Sep 22 '18 at 9:56
  • $\begingroup$ Actually $X \setminus (X \setminus A)=A$. So what do you want? you need counterexamples of clopen sets? $\endgroup$ – Chinnapparaj R Sep 22 '18 at 10:04
  • $\begingroup$ I just can't understand how the set could be clopen and how to show that the set is closed but not clopen. $\endgroup$ – Bogdan Sep 22 '18 at 10:20
  • 3
    $\begingroup$ Which set, specifically, in which topology? $\endgroup$ – Berci Sep 22 '18 at 10:28
1
$\begingroup$

Your first sentence is wrong: what you wrote (not being closed under sequential limit) means the set is not (sequentially) closed, which is not the same as being open. (My topology prof used to say 'a set is not a door.' - because a door is either open or closed, as opposed to a set in a topological space.)

Instead, to prove a set $A$ is open, we need to find a basic open set $U$ (given by a 'basis' for the topology) for each $a\in A$ such that $a\in U\subseteq A$.

The basic open sets in a Euclidean space are the open balls $B_r(x)=\{v:\|v-x\|<r\}$.

In particular in $\Bbb R$ these are the open intervals, and for instance, $\Bbb Q$ is neither open nor closed.

When a nontrivial clopen set $A$ exists in a top.space $X$, the space is called disconnected (we have 2 complementary, nonempty open sets, $A$ and $\bar A=X\setminus A$).
For example, consider $X:=(-1,1)\setminus\{0\}$ with the usual topology (inherited from $\Bbb R$), it's the union of 2 nonempty open sets $(-1,0)$ and $(0,1)$, hence both are clopen in $X$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.