# does Bd(A) = Bd(Cl(A))?

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

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

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).