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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$X$ a separable metric space with no isolated points. If $G \subset X$ is a countable dense $G_{\delta}$ subset of $X$, why is $X$ meager?

Let $X$ be a separable metric space with no isolated points. Then if $G \subset X$ is a countable dense $G_{\delta}$ subset of $X$, why is $X$ meager? I've written a couple things so far, Let ...
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1answer
225 views

Diophantine number has full measure but is meager

This an exercise 3 on Terence Tao's blog: A real number $x$ is Diophantine if for every $\varepsilon > 0$ there exists $c_\varepsilon > 0$ such that $|x - \frac{a}{q}| \geq ...
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2answers
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Examples of closed subspaces of Baire spaces that fail to be Baire?

I am looking for some nice examples of Baire spaces containing closed subspaces that fail to be Baire. Clearly, $X$ should not satisfy either one of the standard hypotheses for the Baire category ...
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1answer
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$X$ is a complete metric space, $Y$ is compact. $X \times Y$ is Baire?

Requesting a hint or solution. X is a complete metric space and Y is a compact hausdorff space. Trying to show that $X \times Y$ is a Baire space.
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1answer
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Can a sequence of functions have infinity as limit exactly at rationals?

Someone asked me this question. And he said it's an exercise from Rudin's Real and Complex Analysis. Does there exist a sequence of continuous functions $f_n(x)$, such that $\lim_{n \to \infty} ...
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1answer
227 views

Are function spaces Baire?

Let $X$ and $Y$ be manifolds and suppose that $X$ is a compact, complete metric space and $Y$ is a complete metric space. So, both $X$ and $Y$ are Baire spaces. Question: For what values of $k\geq ...
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1answer
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Slight generalisation of the Baire category theorem?

By the Baire category theorem, one cannot write a complete metric space $X$ as a countable union of closed nowhere dense subsets of $X$. Can this be generalised to say that there is no injection $f: X ...
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1answer
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Why is there an element with infinite spectrum in a commutative Banach algebra with infinitely many characters?

Let $A$ be a commutative Banach algebra such that set of all characters is infinite. I want to prove that there exists an element in $A$ such that its spectrum is infinite.
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Baire's theorem from a point of view of measure theory

According to Baire's theorem, for each countable collection of open dense subsets of $[0,1]$, their intersection $A$ is dense. Are we able to say something about the Lebegue's measure of $A$? Must it ...
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0answers
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Partition of $\mathbb{R}$ into nullset and 1st category set

Let $\{a_i\}_{i=1}^{\infty}$ be an enumeration of the rationals, $\mathbb{Q}$. Let $I_{ij}$ be the open interval centered at $a_i$ and having length $1/2^{i+j}$, and define $G_j = \cup_{i=1}^\infty ...
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1answer
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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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0answers
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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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2answers
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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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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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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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1answer
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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
263 views

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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1answer
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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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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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1answer
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Is the union of two nowhere dense sets nowhere dense?

Is the union of two nowhere dense sets nowhere dense? Using the following definition: A nowhere dense set is a subset $E\subset X$ of a metric space (or topological space) $X$ such that ...
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2answers
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Baby Rudin problem 3.22: prove Baire's theorem. Am I going in a reasonable direction?

Suppose $X$ is a nonempty complete metric space, and $\{G_n\}$ is a sequence of dense open subsets of $X$. Prove Baire's theorem, namely, that $\bigcap_1^\infty G_n$ is not empty. Hint: find a ...
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1answer
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Let $X$ be an infinite dimensional Banach space. Prove that every basis of X is uncountable.

Let $X$ be an infinite dimensional Banach space. Prove that every basis of $X$ is uncountable. Can anyone help how can I solve the above problem?
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3answers
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A question on the proof of Open mapping theorem

I was following the proof of the Open Mapping Theorem in functional analysis in Wikipedia, and I came across a line in the proof that did not make sense. Some notations: $U,V$ are open unit balls in ...
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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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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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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
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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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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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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
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(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
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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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18answers
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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 ...
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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
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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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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
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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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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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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
133 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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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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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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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 ...
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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 ...
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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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3answers
536 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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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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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?