# Thread: A Property of Boundary of a Set in a Topological Space

1. ## A Property of Boundary of a Set in a Topological Space

Dear Colleagues,

I want a help for proving the following property:

Bd[Bd{Bd(A)}]=Bd[Bd(A)], where Bd denotes the boundary and A is subset of a toplogical space X.

Actually I proved that Bd[Bd{Bd(A)}] is a subset of Bd[Bd(A)] but I could not prove the converse.

Regards,
Raed.

2. ## Re: A Property of Boundary of a Set in a Topological Space

Originally Posted by raed
Bd[Bd{Bd(A)}]=Bd[Bd(A)], where Bd denotes the boundary and A is subset of a toplogical space X.
Have a look at the web reference
Note that $\partial (A)$ is the boundary of A which is a closed set so that $\partial \partial (A) = \partial \partial \partial (A)$

3. ## Re: A Property of Boundary of a Set in a Topological Space

Originally Posted by Plato
Have a look at the web reference
Note that $\partial (A)$ is the boundary of A which is a closed set so that $\partial \partial (A) = \partial \partial \partial (A)$
Thank you for your reply actually using that the boundary is closed I proved one inclusion it is not clear for me how to get the converse.

Regards

4. ## Re: A Property of Boundary of a Set in a Topological Space

Originally Posted by raed
Thank you for your reply actually using that the boundary is closed I proved one inclusion it is not clear for me how to get the converse.
Actually I proved that Bd[Bd{Bd(A)}] is a subset of Bd[Bd(A)] but I could not prove the converse.
Well I have not have any idea what to tell you.
It is clear on that webpage I gave you.
It says "For any set S, ∂S ⊇ ∂∂S, with equality holding if and only if the boundary of S has no interior points, which will be the case for example if S is either closed or open. Since the boundary of a set is closed, ∂∂S = ∂∂∂S for any set S.

If $M$ is a set then the closure $\overline{M}=\text{int}(M)\cup\partial(M)$.
Thus $\partial(M)\subseteq M$ if $M$ is closed.

Is it possible that $\partial\partial(A)$ has any interior points?