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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?

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  • $\begingroup$ Regarding typing math symbols, see here. You can use the "edit" button to update your post with LaTeX formatting. $\endgroup$
    – Nate Eldredge
    Mar 23, 2014 at 14:48
  • $\begingroup$ In addition to Nate's suggestion, you should add more line breaks. Reading a continuous wall of text is very hard. $\endgroup$
    – tomasz
    Mar 23, 2014 at 14:53
  • $\begingroup$ @Nate: If you write [edit] in the comment it compiles to the edit links, like this edit. $\endgroup$
    – Asaf Karagila ♦
    Mar 23, 2014 at 15:08
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    $\begingroup$ @AsafKaragila: Ooh, neat! Thanks. $\endgroup$
    – Nate Eldredge
    Mar 23, 2014 at 15:10

1 Answer 1

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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 $B$.

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  • $\begingroup$ Thanks I think this answers my second question but how can I prove the first one? That is, show that the countable union of En has empty interior using baire category theorem to the dense open sets X\En. $\endgroup$
    – nomadicmathematician
    Mar 23, 2014 at 15:25
  • $\begingroup$ oh wait i made a typo. The text says that En is a sequence of nowhere dense subsets of a complete metric space X. And Baire applies to a sequence of dense open subsets of X. I think that's why the text says apply Baire to the dense open sets X\cl(En) and not X\En this is why I'm stuck. $\endgroup$
    – nomadicmathematician
    Mar 23, 2014 at 15:31
  • $\begingroup$ My guess is that since X\cl(En) is a subset of X\En, the subset order is preserved when you take a countable intersection of either sides and since the cl(⋂n(X∖cl(En))) is X so is cl(⋂n(X∖En)) since closure preserves subset order. Is this proof correct? $\endgroup$
    – nomadicmathematician
    Mar 23, 2014 at 15:34
  • $\begingroup$ @user135204: Yes, you've got it. Baire says that $\bigcap_n (X \setminus \bar{E}_n)$ is dense. By De Morgan, $\bigcap_n (X \setminus \bar{E}_n) = X \setminus \bigcup_n \bar{E}_n$, so by our claim above, $\bigcup_n \bar{E}_n$ has empty interior. Now $\bigcup_n E_n \subset \bigcup_n \bar{E}_n$ so if the latter has empty interior, so must the former. $\endgroup$
    – Nate Eldredge
    Mar 23, 2014 at 15:35
  • $\begingroup$ thanks a lot you've cleared things up for me. By the way I'm guessing that the claim above does not require X be complete and applies to general topological spaces am I correct? $\endgroup$
    – nomadicmathematician
    Mar 23, 2014 at 15:38

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