corollary to baire category theorem

Question

I'm studying topology with gamelin and greene's text and I came across a corollary to the baire category theorem which states that

"Let (En) be a sequence of nowhere dense subsets of a complete metric space X. Then the countable union En has empty interior."

The book merely says apply the baire category theorem to the dense open sets Un=X\cl(En). Applying this, I get that the countable intersection of X\cl(En) is dense. I tried to prove this using the identity int(A)=X(cl(X\A)).

So applying this I figured out that showing that the countable union of En has empty interior is equal to showing that the countable intersection of X\En is dense. I see a similarity of this with the previous result applying the baire category thm but I'm stuck here. How can I prove this?

Also, the book says that the baire category thm is equivalent to the following statement "In a complete metric space X, the complement of a set of the first category is dense in X."

But the complement of a set of the first category say the countable union of En, which is a sequence of nowhere dense subsets would be the countable union of X\En while as the previous corollary suggested Baire category thm applies to the dense ope ts X\cl(En).

How can these two statements turn out to be equivalent?

Answer 1

If I understand your question correctly, you essentially want a proof of the following claim:

A set $A \subset X$ has empty interior if and only if its complement $X \setminus A$ is dense in $X$.

This is quite simple if you look at it the right way. As a hint, it may help to first show or recall that:

A set $B \subset X$ is dense in $X$ if and only if every nonempty open subset of $X$ contains a point of $B$.