Baire Category Theorem & The Real Numbers

604 Views Asked by At

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 :)

2

There are 2 best solutions below

0
On BEST ANSWER

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 $(-\infty,0)\cup(0,\infty)$.

0
On

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.