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.

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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 ...
3
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1answer
137 views

Baire Category and F sigmas

Can anyone give an example of a Meagre subset of $\mathbb{R}$ (i.e., of the first Baire Category), which isn't an $F_\sigma$? Simple cardinality arguments show that such a thing exists, but I can't ...
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Definitions of Baire first and second category sets

From Planetmath A meager or Baire first category set in a topological space is one which is a countable union of nowhere dense sets. A Baire second category set is one which contains a ...
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2answers
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Second category but not locally residual

Let $X$ be a Baire space. A subset $E\subset X$ is said to be of first category if it can be expressed as the union of countably many nowhere dense subsets of $X$. Then $E$ is said to be of second ...
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2answers
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$\lim_{n\to \infty}f(nx)=0$ implies $\lim_{x\to \infty}f(x)=0$

Can anyone help me with this problem? Let $f:[0,\infty)\longrightarrow \mathbb R$ be a continuous function such that for each $x>0$, we have $\lim_{n\to \infty}f(nx)=0$. Then prove that ...
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1answer
324 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 ...
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0answers
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x-section of closure of E of first category implies x-section of E nowhere dense

Let $E$ be a subset of first category of product space $X \times Y$. Why is the following true: if $(\bar E)_x \subset Y$ is of first category then it follows that $E_x$ is nowhere dense. $E_x$ ...
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1answer
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A classical problem about limit of continuous function at infinity and its connection with Baire Category Theorem

When I google "baire category theorem", I get a link to Ben Green's website. And at the end of the paper, he mentioned such a classic problem: Suppose that $f:\mathbb{R}^+\to\mathbb{R}^+$ is a ...
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2answers
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Doubt in Kechris's Classical Descriptive Set Theory

In Theorem 8.29 in the Kechris's book Classical Descriptive Set Theory, he writes that if $W=\bigcup_{i\in I} U_i$ where $U_i$ are pairwise disjoint and set $A$ is comeager in each $U_i$, then $A$ is ...
8
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2answers
710 views

Proof That $\mathbb{R} \setminus \mathbb{Q}$ Is Not an $F_{\sigma}$ Set

I am trying to prove that the set of irrational numbers $\mathbb{R} \setminus \mathbb{Q}$ is not an $F_{\sigma}$ set. Here's my attempt: Assume that indeed $\mathbb{R} \setminus \mathbb{Q}$ is an ...
8
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1answer
192 views

Help understanding game version of Baire category theorem

I got this from Thomson et al.'s freely available "Elementary Real Analysis" p.356. They introduce Baire's category theorem through a game where, given two players (A) and (B) Player (A) is given ...
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2answers
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Corollary to Baire's Category Theorem

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 ...
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5answers
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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 ...
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5answers
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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?
2
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1answer
192 views

disjoint union of Baire spaces which is a Baire space

Say we have a family {$A_\alpha$} of disjoint Baire spaces. Also suppose that each $A_\alpha$ is disjoint from the closure of the union of the other sets. Show that $\bigcup_{\alpha} A_\alpha$ is a ...