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I am taking a Real Analysis unit at University and the topic of the Baire Category Theorem is prevalent in all of the course, however I'm actually, embarrassing stuck right at the start of the definition with the most obvious Complete space (the reals)

The first formulation of the Baire Category Theorem that we have been given is as follows:

Let $(X,d)$ be a complete metric space. If $(G_n)_{n\in\mathbb{N}}$ is a sequence of dense, open sets of X then $$\bigcap_{n=1}^\infty G_n $$ is dense in $X$.

Anyway, my question is, Consider $\mathbb{R}$ as the metric space. I can think of the $G_n$'s as perhaps, the rationals and irrationals. These are both dense, and the reals is complete. However their intersection is clearly the empty set, which is nowhere dense in $\mathbb{R}$. Surely this breaks the theorem right away? Or am I just being stupid!

Thanks for your help :)

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    $\begingroup$ Yes but neither the rationals nor the irrationals are open sets. $\endgroup$ – Gregory Grant Jan 2 '16 at 11:42
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    $\begingroup$ No problem, it's a common mistake most people make at least once. $\endgroup$ – Gregory Grant Jan 2 '16 at 11:45
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    $\begingroup$ If you can accept my answer below I'd appreciate it. We work for upvotes :-) $\endgroup$ – Gregory Grant Jan 2 '16 at 11:47
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    $\begingroup$ @user291103 I've noticed you've yet to accept an answer for your questions on here. You can do so by clicking on the check mark below the downvote arrow on the answer you would like to accept. It will help boost your rep on the website and give you more privileges :) $\endgroup$ – Brandon Thomas Van Over Jan 2 '16 at 12:07
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    $\begingroup$ A good example of a sequence of sets to which we can apply the theorem is the following $\{\mathbb R\setminus\{x\}: x\in\mathbb Q\}$. Their intersection is precisely the irrationals, which is dense $\endgroup$ – SamM Jan 2 '16 at 13:21
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Yes but neither the rationals nor the irrationals are open sets. Open dense sets look more like $\Bbb R\setminus\Bbb Z$. Or even more simply $(-\infty,0)\cup(0,\infty)$.

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The theorem states that this holds for open dense sets. While both of the subsets you chose are dense, neither is open, so this does not meet the criteria of the theorem.

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