Give a counterexample disproving: $Boundary(Closure(A)) = Boundary(A)$

Question

I know that for set $A$, Closure($A$) = $A \cup L$ where $L$ is the set of limit points.

So $Boundary(A \cup L) = Boundary(A) $

It doesn't seem like it's true necessarily I just cannot give a concrete example.

Could anyone give an example showing A and it's closure, and its set of limit points please.

Answer 1

$A = \mathbb{Q}$ in the reals, usual topology.

its closure is $\mathbb{R}$ which has empty boundary.

its boundary is $\mathbb{R}$ which is non-empty.

Answer 2

How about a punctured disk?

Its closure is the whole disk, which has boundary a circle.

But the punctured disk's boundary is the circle along with the point in the middle.