I have been reading this post for awhile and the way you structured the proof does not make any sense to me.

Is there any other way to show this proof?

like with if X is in S or if x is not S sort of version then there exists a deleted neighborhood intersected with S that is nonempty?

I kind of have the jist of what accumulation points, boundary and interior points are but im having trouble forming a proof in my own words.

S is a subset of R.

I know that if x is an accumulation point there exists

S =/=

. So by that if x is in the deleted neighborhood and an accumulation point (meaning x is in S'), then x is not in S.

If x is not in S then x is in R/S.

So if x is in R/S then that means that

=/=

So x in the bd of S.

Is this kind of on the right track?