Let $A$ be subset of a topological space $X$. Show that: $ b(A) \cap A= \emptyset \iff A$ is open.

I started the proof like this: (<==) Let $A$ is open. Then (by a proposition)


Also $b(A)=(int(A) \cup ext(A))^C $

So $b(A) \cap A $ implies $((int(A)\cup ext(A))^C ) \cap A $

implies $((int(A))^C \cap ext(A)^C) \cap int(A)$ (using proposition)

implies $ ((int(A))^C \cap int(A) \cap (ext(A)^C))$

implies $ (\emptyset \cap (ext(A)^C))$

implies $\emptyset$

This is what I have done so far. Is it a right way to prove the implication? And how should I prove the reverse implication?


Let a belong to A then we have suppose a is not in int. A this implies every neighbourhood of a has a point of X-A then we have a is a limit point of X-A thus a belongs to b(A) and A a contradiction. Thus every point is interior and hence A is.open


$x\in\partial A$ if and only if for every neighborhood $U$ of $x$ we have: $$U\cap A\neq\varnothing\text{ and } U\cap A^c\neq\varnothing$$

If $A$ is open we can take $U=A$, to find easily that this is not true.

So we conclude that $x\notin\partial A$ and proved is now that:

$$A\text{ open}\implies A\cap\partial A=\varnothing$$

If conversely $A\cap\partial A=\varnothing$ and $x\in A$ then $x\notin\partial A$.

That means that some neighborhood $U$ of $x$ must exist with: $$U\cap A=\varnothing\text{ or }U\cap A^c=\varnothing$$ We have $x\in U\cap A$ so conclude that $U\cap A^c=\varnothing$ or equivalently $U\subseteq A$.

This can be proved for every $x\in A$ so proved is now that:$$A\cap\partial A=\varnothing\implies A\text{ open}$$


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.