Disprove any closed set has a point on its boundary I am told that the statement "any closed set has a point on its boundary" is false, yet I don't know how to disprove it. In fact, I think it is true. 
Suppose we have [a,b], a closed set. Then, the boundary would be {a,b}, both of which are the elements of the set. So, there we have a closed set that has point(s) on its boundary.
I think one reason why I was told that the statement is false can be related to the empty set since


*

*Empty set is also closed (clopen, but still)

*It does not have a point on its boundary

*The statement goes "any closed set..", hence empty set is not
excluded.


Supposing that the statement is false due to the above-mentioned reasoning (fact that empty set was not excluded), would "any nonempty closed set has a point on its boundary" be correct?
Thanks.
 A: Let $X$ be a topological space and $C\subseteq X$ a closed set. Then $cl(A)=A$ (closure of $A$ is equal to $A$). Our set $A$ has empty boundary if
$$ \emptyset = \partial A = cl(A) \setminus int(A).$$
As $int(A) \subseteq clo(A)$ this is equivalent to $int(A)=cl(A)$. This means, the closed sets with empty boundary are exactly the clopen sets. 
Thus, if $X$ is connected, then there are only two sets, which are closed and have empty boundary, namely $\emptyset$ and $X$.
A: It depends whether you are talking about general topological spaces or just $\mathbb{R}$. In $\mathbb{R}$ there are indeed very few examples: the empty set, as you mentioned, and all of $\mathbb{R}$.
In general topological spaces, however, you may have more examples.
It's not hard to see a set is closed if and only if it contains its boundary, and is open if and only if its intersection with its boundary is empty. Therefore, a set has an empty boundary if and only if it's clopen. Therefore in any connected space (like $\mathbb{R}$) the only closed sets satisfying this are the whole space and the empty set, but in disconnected spaces you will have more examples.
