Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In Rudin's Real and Complex Analysis (p. 97 in my 3rd edition), the following is stated as a corollary to Baire's Category Theorem:

"In a complete metric space, the intersection of any countable collection of dense $G_{\delta}$'s is again a dense $G_{\delta}$."

Proof: "This follows from the theorem, since every $G_{\delta}$ is the intersection of a countable collection of open sets, and since the union of countably many countable sets is countable."

I don't understand the word union in the argument. It seems like it should be intersection, as the theorem states that such an intersection is dense. Is this a typo or am I missing something?

share|improve this question
1  
Union (as in set union) is the correct word. Assuming the theorem, we just need to show that a countable collection of $G_\delta$'s is again countable, therefore satisfying the assumptions of the theorem. –  jericson Jan 4 '11 at 23:11

2 Answers 2

up vote 7 down vote accepted

"Union" is the correct word here. The point is that we have an iterated intersection: we have for each $i$ a countable family $\{U_{ij}\}$ of dense open subsets and we are taking the intersection over all $i$: $\bigcap_{i \in I} \bigcap_{j \in J} U_{ij}$.
Here $I$ and $J$ are both countable index sets. This intersection can be written as $\bigcap_{(i,j) \in I \times J} U_{ij}$. The passage you quote is giving an argument that $I \times J$ is again a countable index set: it is the countable union of the countable sets $\{i\} \times J$ as $i$ ranges over all elements of $I$. (One could give other arguments for this as well.)

share|improve this answer
    
Ah! So we're basically throwing all of the open sets together and showing that we have only countably many (thus satisfying the hypothesis). I understand now, but I still would argue with the choice of the word union here! –  dls Jan 4 '11 at 23:21
    
So the union is really over the indices, right? –  dls Jan 4 '11 at 23:36
    
Yes. $ \textbf{} $ –  Pete L. Clark Jan 5 '11 at 0:19
1  
@dls: The reason why "union" is the right word here is because the Baire category theorem is about (countable) collections of some sets (open, in this case), and to apply the theorem, you have to take the union of several such collections to form the appropriate one to which the theorem applies. This is a common situation in mathematics, you put together several collections into a single one (by taking their union). In this case, what you do with the sets in each collection is to intersect them, but that is secondary. –  Andres Caicedo Jan 5 '11 at 2:17

Actually, when I first read the quote, I just assumed Rudin was talking about what happens when you take complements. And that might actually be an easier way to see what's going on.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.