Proof that the empty set is closed I just want to confirm if my proof is correct for the statement: "The empty set is closed". Although I know that the empty set is also open, the proof is similar.
Proof. Suppose $∅$ is open, then for all x that is an element of $∅$ there exist a ball $B(x,r)$ that is centered at x with radius r such that $B(x,r)$ is a subset of $∅$. But this implies that $∅$ is not the empty set since it contains some elements. Thus, if $∅$ is open then $∅$ is not the empty set, but arguing contrapositively, if $∅$ is the empty set then $∅$ is closed. QED
 A: Using your method to 'prove' that $\varnothing$ is open and closed you should reason like this:


*

*Is it true that for every element $x$ in $\varnothing$ there is an open ball $B(x,r)$ with $x\in B(x,r)\subseteq\varnothing$? Yes! Because there is no element in $\varnothing$ for wich this is not true. Even stronger: there is no element in $\varnothing$ at all! Conclusion: $\varnothing$ is open.

*Is it true that for every element $x\in X$  there is an open ball $B(x,r)$ with $x\in B(x,r)\subseteq X$? Yes! That is obvious. We conclude that $X$ is open or equivalent that its complement $\varnothing$ is closed.
A: The empty set is both open and closed. By definition of a topology both the whole space and the empty set are open. Since the empty set is the complement of the whole set it is also closed. Your proof does not work since the condition for being an open set you want to use states that a set $A$ is open if  any $x$ in $A$ is contained in a basic open set contained in $A$. But since the empty set has no elements, there is nothing to check and the condition is trivially satisfied.
A: No, your argument is flawed. 
When you say that it contains some elements, then that's where you get wrong. If you said for all elements $x$ then $P$ where $P$ is some property, then that doesn't imply that it has an element. It's an example of what is called a vacuously true statement. If you want to argue it, you can prove that its negation is false. This is how it would go. Assume there is $x$ such that $\not\  P$ , but there is no $x$, so the negation is false. Thus, it is true.
Now, you can argue in the same manner to prove that it's closed.
Assume it contains a limit point that doesn't belong to it. But it has no limit points, then it doesn't contain any limit point which doesn't belong to it. Then, its all limit points (which are non-existent basically) belong to it, and we are done.
