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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A property dealing with complete metric spaces

I came across a property in a textbook that caught my eye. The property is: If $X$ is a complete metric space, then the intersection of any two dense $G_{\delta}$-subsets of $X$ is dense in $X$. This ...
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1answer
170 views

(ZF) Equivalent statements to Baire Category Theorem

So far, I have proved following two for a polish space $X$; 1.If $\{F_n\}$ is a family of closed subset of $X$, where $X=\bigcup_{n\in \omega} F_n$, then at least one $F_n$ has a nonempty inteior. ...
3
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1answer
277 views

Baire Category Theorem

This is Asaf's argument; (ZF) If $\mathbb{R}^k$ is a countable union of closed sets, then at least one has a nonempty interior Suppose that $(X,d)$ is a separable complete metric space, and ...
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4answers
157 views

Complement of co-dense set.

Asaf's argument : (ZF) If $\mathbb{R}^k$ is a countable union of closed sets, then at least one has a nonempty interior Let $X$ be a separable complete metric space. Let $D$ be a countable debse ...
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4answers
309 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 $\mathbb{R}^k$. ...
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2answers
178 views

$x$-axis is meager set on $\mathbb{R}^2$

Subset $A$ of metric space $X$ is meager on $X$, iff $\text{IntCl}A=\emptyset$. But, why $x$-axis is meager set on $\mathbb{R}^2$? My attempt (please don't kill me): ...
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19answers
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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 ...
105
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1answer
3k 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 ...
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1answer
176 views

Question missing condition in Royden Exercise 7.42 b, about Baire Category

In Royden's Real Analysis P164 Q7.42b, It assumes that $X$ and $Y$ are complete metric spaces. Let $O$ be a dense open set in $X \times Y$. Assertion: Then there is a $G \subset X$ which is a ...
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2answers
259 views

Example of Baire Space

Can anybody supply an example of a Baire Space, that is neither locally compact nor metrizable. I would be gratefull also for some references.
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1answer
359 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 ...
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1answer
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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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2answers
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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
142 views

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
606 views

$\lim_{n\to \infty}f(nx)=0$ 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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2answers
371 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
393 views

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
785 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
198 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
556 views

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?
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1answer
196 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 ...