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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Use the concept of Baire Category to prove that the upper integral of a positive function is always positive.

Use Baire Category to prove that the upper integral from $0$ to $1$ of any $f:[0,1]\to(0,1]$ is greater than $0$.
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Baire's theorem [duplicate]

I want to know interesting applications of Baire's Category Theorem. For example existence of no where differentiable function. Can any body tell me some similar applications?
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1answer
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Complement of a meagre subset of $\mathbb{R}$ contains an uncountable $G_\delta$ set

I'm trying to show that the complement of a meagre set $A \subseteq \mathbb{R}$ contains an uncountable $G_\delta$ set. Here is what I got so far : since $A$ is meagre, there exists nowhere dense ...
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$\mathbb{Q}$ is not locally compact using baire category?

is there any result related together with Baire Category Theorem, Locally compactness, and Completeness? actually I would like to prove $\mathbb{Q}$ is not locally compact. I realize that singleton ...
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About Baire's Category Theorem(BCT)

Consider the following theorem known as Baire's Category Theorem (BCT). Theorem.[BCT] A non-empty complete metric space $X$ is not a countable union of nowhere dense sets. I am interested on how to ...
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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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1answer
251 views

The Principle of Condensation of Singularities

Let $X$, $Y$ be Banach spaces and $\{T_{jk} : j,k \in\Bbb N\}$ be bounded linear maps from $X$ to $Y$. Suppose that for each $k$ there exists $x\in X$ such that $\sup\{\lVert T_{jk} x\rVert : j ...
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2answers
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Cantor's intersection theorem and Baire Category Theorem

From an old post in math stackexchange, I read a comment which goes as follows " I like to think of Baire Category Theorem as spiced up version of Cantor's Intersection Theorem". My question -----is ...
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3answers
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Dirichlet Function Pointwise Convergence

We say that a function $f$ is Baire Class $1$ if there is a sequence of functions $f_i \to f$ pointwise where each $f_i$ is continuous. The set of discontinuities of a Baire Class $1$ function $f$ ...
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Sets of second category-topology

A set is of first category if it is the union of nowhere dense sets and otherwise it is of second category. How can we prove that irrational numbers are of second category and the rationals are of of ...
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Dense subset of Cantor set homeomorphic to the Baire space

Does anyone know a proof that the Cantor set, $\{0,1\}^{\mathbb{N}}$, has a dense subset homeomorphic to the Baire space, $\mathbb{N}^{\mathbb{N}}$? Thank you.
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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 ...
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1answer
369 views

Baire: Show that $f\colon \mathbb{R}\to\mathbb{R}$ is a polynomial in an open bounded set

Let $f\colon\mathbb{R}\to\mathbb{R}$ be an infinitely differentiable function, and suppose that for each $x\in\mathbb{R},$ $\exists\ n=n(x)\in\mathbb{N}$ such that $f^{(n)}(x)=0$. For each fixed ...
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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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1answer
166 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.
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1answer
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Is $\ell^1 \subset \ell^2$ meagre? [duplicate]

Possible Duplicate: Prove $\ell_1$ is first category in $\ell_2$ Consider $\ell^2$ with the topology induced by the usual norm. We can easily prove that $\ell^1 \subset \ell^2$. I am ...
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1answer
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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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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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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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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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1answer
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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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1answer
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(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. ...
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1answer
289 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
327 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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1answer
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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
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$x$-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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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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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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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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1answer
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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 ...
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1answer
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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 ...
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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 ...