This tag is intended for questions on topics related to Baire category, such as Baire category theorem, meager sets (set of first category), nonmeager sets (set of second category), Baire spaces etc.
50
votes
16answers
2k views
Your favourite application of the Baire category theorem
I think I remember reading somewhere that the Baire category theorem is supposedly quite powerful. Whether that is true or not, it's my favourite theorem (so far) and I'd love to see some applications ...
34
votes
5answers
1k views
Why is it that $\mathbb{Q}$ cannot be homeomorphic to _any_ complete metric space?
Why is it that $\mathbb{Q}$ cannot be homeomorphic to any complete metric space?
Certainly $\mathbb{Q}$ is not a complete metric space. But completeness is not a topological invariant, so why is the ...
4
votes
1answer
115 views
Prove $\ell_1$ is first category in $\ell_2$
Prove that $\ell_1$ is first category in $\ell_2$.
I tried to solve this, but had no idea about the approach. Any suggestions are helpful.
Thanks in advance.
4
votes
4answers
199 views
(ZF) If $\mathbb{R}^k$ is a countable union of closed sets, then at least one has a nonempty interior
Since the specific space $\mathbb{R}^k$ is given, this might be provable in ZF.
Let $\{F_n\}_{n\in \omega}$ be a family of closed subset of $\mathbb{R}^k$, of which the union is $\mathbb{R}^k$.
...
4
votes
1answer
186 views
Application of Baire category theorem in Moore plane
The proof that Moore plane is not normal I have read was using Cantor's nesting theorem.
But I heard that it is also possible to use Baire category theorem to prove and I want to know how.
So, as ...
84
votes
1answer
2k views
Does the open mapping theorem imply the Baire category theorem?
A nice observation by C.E. Blair1, 2, 3 shows that the Baire category theorem for complete metric spaces is equivalent to the axiom of (countable) dependent choice.
On the other hand, the three ...
33
votes
5answers
3k views
Is $[0,1]$ a countable disjoint union of closed sets?
Can you express $[0,1]$ as a countable disjoint union of closed sets, other than the trivial way of doing this?
6
votes
1answer
252 views
Every space is “almost” Baire?
There is this theorem called the Banach category theorem which states that in every topological space any union of open sets of first category is of first category.
Now doesn't this imply that every ...
6
votes
1answer
306 views
Infinitely times differentiable function
Let $f$ belong to $ C^{\infty}[0,1]$ and for each $x \in [0,1]$ there exists $n \in \mathbb{N}$ so that
$f^{(n)}(x)=0$. Prove that $f$ is a polynomial in $[0,1]$.
I am trying to use Baire Category ...
