0
$\begingroup$

prove or disprove

Let (X, T ) be a topological space and let A ⊆ X. Then Bd(A) = Bd(Cl(A)).

I think it is false statement ,could you help me with counterexample please

$\endgroup$
  • $\begingroup$ sorry, I mean Bd(A) = Bd(Cl(A)) $\endgroup$ – rian asd Apr 25 '17 at 3:28
  • $\begingroup$ Be careful next time, what you wrote earlier was true, this is likely to be false. $\endgroup$ – астон вілла олоф мэллбэрг Apr 25 '17 at 3:32
  • $\begingroup$ yeah it is false but I do not know how can I find counterexample $\endgroup$ – rian asd Apr 25 '17 at 3:40
  • $\begingroup$ Do we have Cl(A)=A U Bd(A) ? $\endgroup$ – Jacob Wakem Apr 25 '17 at 4:39
  • $\begingroup$ I am not sure what you mean exactly $\endgroup$ – rian asd Apr 25 '17 at 4:42
2
$\begingroup$

We know that $Bd(A) = \overline A - Int(A)$

So, $Bd(\overline A) = \overline {\overline A} - Int(\overline A)$.

Clearly, the two will not be equal if $Int(A) \neq Int(\overline A)$.

This is possible if $A = \mathbb Q$, for then $Int(A) = \emptyset$ (every rational has an irrational as close as you like), while $\overline(A) = \mathbb R$, so it's interior is the entire real line (which is of course not the empty set).

Hence, the proposition is false.

$\endgroup$

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.