0
$\begingroup$

I know that the boundary of a set $A \subset X$ is closed. Is it also true that a subset of a boundary is closed?

Thank you very much for your help

$\endgroup$
  • 4
    $\begingroup$ No. Is every subset of a closed set closed? $\endgroup$ – user296602 Mar 13 '16 at 21:33
  • $\begingroup$ @T.Bongers Not every closed set is a boundary. $\endgroup$ – egreg Mar 13 '16 at 21:35
  • $\begingroup$ No, I know it is not true in the general case, just thought if there was an exception in the case of a boundary. $\endgroup$ – stensootla Mar 13 '16 at 21:35
  • $\begingroup$ @egreg Indeed, but it's a good starting point to address whether it's reasonable to expect a yes answer. $\endgroup$ – user296602 Mar 13 '16 at 21:37
  • $\begingroup$ Why would it be? subsets of closed sets aren't closed in general, so why would the set being a boundary j force its subsets to be closed. Consider A={(x,y)|y $le $ 2}. Its boundary is the line y=2. This line will have a subset (0,1)x {2}. Which is not closed. Consider any boundary A with more than one point and at least on limit point,a. Consider A/a. That can't be closed. $\endgroup$ – fleablood Mar 13 '16 at 22:28
3
$\begingroup$

No: $$\mathbb{Q} \subseteq \partial \mathbb{Q} = \mathbb{R}.$$

$\endgroup$
1
$\begingroup$

No. Consider the set $A=\{(x,y):x^2+y^2<1\}$ in two-space and the subset of the boundary $$ T=\{(x,y):x^2+y^2=1,y>0\} $$ This set is not closed, because $(1,0)$ belongs to the closure, but not to $T$.

$\endgroup$
1
$\begingroup$

Let $A$ be a boundary with at least two points and at least one limit point. (Finding such a boundary is trivial. The unit circle in Euclidean plane will do.) Let $a $ be one of the limit points. Then $a$ is a limit point of the set A-{$a $}. So the set $A-{a}\subset A $ is not closed.

Closed subsets, in general (not always, of course-- sets of isolated points, which have no limit points would be an exception), will usually have subsets that are not closed. Boundaries are no different.

$\endgroup$

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.