Every interior point of 'the closure of S' is in Int S?

[pantin]

Homework Statement

Let S be a set in R^n, is it true that every interior point of 'the closure of S' is in Int S? Justify.

2. Relevant theorem

S^int = {x belongs to S: B(r,x) belongs to S for some r>0}
The closure of S is the union of S and all its bdary points.

The Attempt at a Solution

My answer is yes, but I am not sure how to give a proof, anyone can give a counterexample?

 

[jostpuur]

It can happen that $$\textrm{int}(S)=\emptyset$$ while on the other hand $$\textrm{int}(\overline{S})\neq\emptyset$$.

 

[pantin]

It can happen that $$\textrm{int}(S)=\emptyset$$ while on the other hand $$\textrm{int}(\overline{S})\neq\emptyset$$.

hi jostpuur, thanks for your reply, but when the first case happens, how can the second one happens like that?
I don't quite get it, can you explain it a little bit more?

 

[jostpuur]

If S doesn't have interior, it doesn't necessarily mean that S is somehow "thin" (like n-k dimensional manifold in n dimensional space, with k > 0), but S can also be "filling" all n dimensions in the space, but on the other hand being "infinitely full of holes" so that interior doesn't exist.