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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Prove that if B is closed and nowhere dense, $f^{-1}(B)$ is also closed and nowhere dense.

Let $f: \mathbb R \rightarrow \mathbb R$ be a continuous function that is nonconstant on any interval. Prove that if B is closed and nowhere dense, $f^{-1}(B)$ is also closed and nowhere dense. My ...
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Why need open in the Baire Category Theorem

In the statement of the Baire Category Theorem, one needs to include openness of a set so that the theorem holds. Question: what is the example such that countable intersection of dense sets is not ...
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1answer
47 views

Baire category of subset $\mathbb{R} \setminus B$ with lebesgue zero measure $B$

Let $\mathbb{Q}=\{ q_i\}^{\infty}_{i=1}$. Let $I_{ij} = \left(q_i - \delta_{ij}, q_i + \delta_{ij}\right)$, $\delta_{ij}=2^{-i-j-1}$ for all $i,j \geq 1$ be an open interval. For every $j\geq1$, ...
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1answer
69 views

Baire's Theorem and Irrationals

I am asked to show that the irrational numbers are not a countable union of closed subsets of $\mathbb{R}$ given that if a complete metric space is the countable union of of closed subsets then at ...
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Outline of Generic Separable Banach Spaces don't have a Schauder Basis

So, I know P. Enflo showed that there is a separable Banach Space that doesn't satisfy the approximation property. My professor mentioned during class that in fact generic separable Banach Spaces don'...
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1answer
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Corollary of Baire theorem

Can anybody help me to prove the following result? Corollary of Baire theorem: Let $(K_j)_{j>0}$ be an increasing sequence of compact sets in $C^n$ and $X$ a bounded open set such that $\overline ...
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Proving a result in Baire categories

If $A\subset X$ is a $G_\delta$ set and dense in $X$, then what I want to show is that $X\setminus A$ is of the first Baire category. Where the meaning of first category is that I can write $$X\...
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1answer
48 views

On the zero set of a $C^2$ function on $[0,1]^2$ [closed]

Let $f:[0,1]^2\rightarrow \mathbb{R}$ be a twice continuously differentiable function with the property that for all $x\in [0,1]$, there is an interval $I_x\subset [0,1]$ such that $f(x,y)=0$ for all $...
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1answer
57 views

continuous an open image of a meager set is meager

I want to know if the following is true. Let $X$ and $Y$ be topological spaces and $f\colon X\to Y$ a continuous open surjection. Suppose that $X$ is meager, then $Y$ is meager. Recall that a meager ...
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1answer
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Question related to Baire category theorem

I met this question from a faculty booklet which I had trouble on involving metric spaces, stating: Let $ (X,d) $ be a complete metric space with the sequence of closed sets, $ \{F_n\}_{n \in N} $ ...
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1answer
56 views

Is $\{1/n : n = 1, 2, 3, …\}$ completely metrizable?

Is $\{1/n : n = 1, 2, 3, ...\}$ with the subspace topology from $\mathbb{R}$ completely metrizable? As as result of Baire's category theorem, we know that if a metric space is complete and there are ...
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1answer
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Locally varying, continuous functions on $R^2$, show $R^2$ cannot be written as $\cup^\infty_{i=1}\cup^\infty_{j=1} \{x : f_i(x) = c_j\}$\}

Problem Statement: If a real valued function on $\mathbb R^2$ is locally varying (on any non-empty open subset $U \subset \mathbb R^2$, the function is not constant), show that $\mathbb R^2$ cannot be ...
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1answer
27 views

Is there an example of a topological space that is of the Second Baire Category but is not a Baire space?

By being of the second category I mean that it is not the countable union of nowhere dense sets and by Baire space I mean a space such that a countable intersection of open dense sets is dense in X. ...
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1answer
34 views

How does this follow from the Baire category theorem?

The book says that statement 2 is a direct consequence of statement 1. I don't see how they prove statement 2 directly from statement 1, can you please help me? Statement 1: A complete metric ...
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1answer
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Show that $R^2$ cannot be written as a countable union of zero sets of non-trivial polynomials

Problem statement: Show that $R^2$ cannot be written as a countable union of zero sets of non-trivial polynomials. Note that the zero set of a polynomial $p(x,y)$ is $\{(x,y) : p(x,y) = 0$}. My ...
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1answer
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How is Baire category theorem used here?

The following is a doubt that arouse from reading this paper by Bandyopadhyay, Jarosz and Rao. Let $E$ be a Banach space and $E^{*}$ be its dual space. Let $e_{0}$ be an element of norm one in $E$ ...
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1answer
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Second Baire class and Borel measurable

I am new to Baire class theory, but need it for one part of a project I am working on. I have seen it referenced that functions of second Baire class are Borel measurable. For example here in this ...
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Lebesgue characterization of Baire class 1 functions

Lebesgue's characterization of Baire class one functions on $\mathbb R$ is the following: $f:\mathbb R \rightarrow \mathbb R$ is Baire class one iff for all $r\in \mathbb R$ $\{ x: f > r \}$ and $...
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1answer
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Application of Baire Category theorem

Suppose that $f$ is infinitely differentiable on $[a,b]$ and suppose that for any $a ≤ x ≤ b$ the Taylor series of $f$ has positive radius of convergence at $x$. Use the Baire Category Theorem to ...
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1answer
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Is $C[0,1]$ equipped with $\lVert \cdot \rVert_1$ a countable union of nowhere dense sets?

Let's consider the space of all continuous function $C[0,1]$ on the intervall $[0,1]$. But instead of using the usual supremum norm we use the $L^1$-Norm $: \lVert f \rVert_1=\int_0^1 \lvert f(x) \...
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1answer
238 views

Dense $G_{\delta}$ set implies comeagre set

Suppose that $X$ is a metric space. Show that if $D$ is a dense $G_{\delta}$ set, then $D$ is comeagre, that is, countable intersection of dense sets. My attempt: Let $D=\bigcap_{n \in \mathbb{N}}{...
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1answer
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Prove: $f: \mathbb{R} \rightarrow \mathbb{R}$ st for every $x \in \mathbb{R}$ there exists $n$ st $f^{(n)}(x) = 0$, f is a polynomial.

If $f: \mathbb{R} \rightarrow \mathbb{R}$ is a smooth function such that for every $x \in \mathbb{R}$ there exists $n$ such that $f^{(n)}(x) = 0$, then f is a polynomial. I'm kind of lost on this ...
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1answer
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The set of continuity of a pointwise limit of continuous functions

Let $\{x_n(t)\}_{n=1}^{\infty}$ be real a sequence of continuous function from $[0,1]$ to $\mathbb{R}$, and $\{x_n(t)\}_{n=1}^{\infty}$ converges pointwise to $x(t)$ i.e. $\lim_{n \to \infty} x_n(t) =...
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2answers
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Undestanding an assumption in Baire's theorem proof.

I was reviewing the proof of Baire's theorem I saw in class a few days ago, and there's an assumption that I didn't managed to see where it was used. I'll put here the proof given. Theorem (Baire) ...
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1answer
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Is the closure of a meager set meager?

How to show that the closure of a meager set is meager? I tried like this: Suppose that it is not meager then $cl(A)$, where $A$ is a meager set in a metric space $(X,d)$ contains an interior point ...
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0answers
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Closed subsets of empty interior are of 1st category

In a metric space is it true that closed sets with empty interior are of 1st category? I.e., that it can be represented as a at most countable union of meager sets? Thanks
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1answer
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Baire Category theorem - Determining First or Second Category

I don't have a mathematician background (I am an engineer). I understand some concepts but the Baire Category theorem and the ideas of first and second category are still very abstract for me. I have ...
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A function continuous on rational points and discontinuous on irrational points [duplicate]

How to find function $f : \Bbb R \to \Bbb R$ such that $f$ is continuous on the rational numbers and discontinuous at irrational numbers? I was told to use the Baire Theorem to show that the set of ...
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1answer
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Prob. 5, Sec. 27 in Munkres' TOPOLOGY, 2nd ed: Every compact Hausdorff space is a Baire space

This is problem 5 in section 27 of Munkres' TOPOLOGY, 2nd ed Let $X$ be a compact Hausdorff space; let $\{A_n\}_{n\in \mathbb{N}}$ be a countable collection of closed sets of $X$. If each set $A_n$ ...
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Hölder continuous functions are of 1st category in $C[0,1]$

I'm trying to show that the Hölder continuous functions in $[0,1]$ are a set of first category in $C[0,1]$. Does it suffice to show that they are not an open subset of $C[0,1]$? Let $\varepsilon>...
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Q: Nowhere dense sets.

Given $X$ a metric space, $A\subset X$ a nowhere dense set. Show that every open ball $B$ contains another open ball $B_1 \subset B$ such that $B_1 \cap A = \emptyset$. EDIT: I modify my proof with ...
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2answers
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Show that a countable dense subset $D \subset X$ is not a $G_{\delta}$

Given $X$ a complete metric space with no isolated points and $D \subset X$ a countable dense subspace, show that $D$ is not a $G_{\delta}$. I am quite lost in trying to use the hypothesis of the ...
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1answer
68 views

Show that $\mathbb{R}^n$ cannot be written as a countable union of proper subspaces

Show that $\mathbb{R}^n$ cannot be written as a countable union of proper subspaces Ok so I know I have to use Baire's Cathegory Theorem here. And I've done the following, lets suppose on the ...
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1answer
97 views

Requirements for the principle of uniform boundedness

The version of the principle of uniform boundedness as we stated it in the lecture seems wrong to me in multiple points. Here is how I would state and proof the principle in the terms we used in the ...
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1answer
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intersection about the second category

$G$ is a locally compact Hausdorff topological group, $A$ and $B$ are two Borel subsets of $G$, and $A$ and $B$ are both of the second category in $G$, then there exist an element $x\in G$, such that ...
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Topological group, which is second category in itself, is a Baire space.

A Baire space is a topological space in which the union of every countable collection of closed sets with empty interior has empty interior. $G$ is a topological group, if $G$ is of the second ...
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1answer
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Sigma-compact Polish groups

I would like to see an example of a sigma-compact Polish group which is not locally compact. I know that e.g. $l^{\infty}$ is a topological group which is sigma-compact but not locally compact. But ...
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1answer
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Baire category theorem in a Banach space

For any two distinct $u,v$ in a countable dense subset of separable real Banach space $X$, let $S(u,v) = \{f \in Y \mid f(u)=f(v)\}$, where $Y$ is the dual space of $X$. Each of $S(u,v)$ is a proper ...
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1answer
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Why this set is of the second category?

I'm watching Baire space on en.wikipedia.org, and find this conclusion. Here is an example of a set of second category in $\mathbb R$ with Lebesgue measure zero. $$\bigcap_{m=1}^\infty \bigcup_{n=1}^...
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1answer
195 views

On the proof of Baire category theorem

I would like to ask about the proof of Baire Category theorem found on wolfram. The excerpt is as below: Baire's category theorem, also known as Baire's theorem and the category theorem, is a result ...
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Producing $\mathbb{R}$ with countable amount of sets?

Prove, that you can't "produce" $\mathbb{R}$ with countable amount of sets, which are nowhere dense(I am not sure I said this definition correct, with nowhere dense, I mean that $Int(\overline X) = \...
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1answer
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If $f:\mathbb R \to \mathbb R$ is an additive function whose graph is $G_{\delta}$ in $\mathbb R^2$ , then the graph is closed in $\mathbb R^2$?

If $f:\mathbb R \to \mathbb R$ is an additive function i.e. $f(x+y)=f(x)+f(y) ,\forall x,y \in \mathbb R $ such that $G(f):\{(x,f(x)) : x\in \mathbb R\}$ is a countable intersection of open sets , ...
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1answer
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Is it true that every 1st category subset of a 2nd category space has empty interior?

Let $X$ be a metric space. Are these conditions equivalent: Each set of the 1. category in $X$ has empty interior; $X$ is of the 2. category. It is obvious that $1 \Rightarrow 2$. Is it true that $...
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1answer
42 views

Why empty set? (consequences of Baire's theorem)

I did not understand the proof of Theorem 5.13 of Rudin, [Real and Complex Analysis]. See next. In a complete metric space X which has no isolated points, no countable dense set is a $G_{\delta}$. ...
3
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1answer
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Question about the Baire space, $\sigma$-algebra and $\sigma$-ideal.

Let $\text{BP}(X)$ denote $\sigma$-algebra of subsets of $X$ with the Baire Property BP and $\text{MGR}(X)$ denote the $\sigma$-ideal of meager sets in $X$. Assume $X$ is second countable Baire space....
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Question about of Baire property and Baire space

In reading Kechris book. Please, I would like help with this proposition. For convencion we put for $A \subseteq X$, $$\sim A=X\setminus A$$ If $A$ is comeager in $U$, we say that $U$ forces $A$, ...
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Proving a linear function is bounded using the Baire category theorem (or its consequences)

This is a problem from Folland. Let $\mathcal{X}, \mathcal{Y}$ be Banach spaces. If $T : \mathcal{X} \rightarrow \mathcal{Y}$ is linear and $f \circ T \in \mathcal{X}^*$ for all $f \in \mathcal{Y}^*,...
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1answer
49 views

Second category subset vs subspace

Let $(X,T)$ be a topological space. When we say $A$ is first category in $(X,T)$ we mean that it is the union of (or is covered by) countably many sets which are nowhere dense in $(X,T)$. $A$ is ...
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Non differentiable solutions to $\partial_x f + \partial_y f =0$

This nice paper by Gilles Godefroy (in French) tells us the story of Baire's lemma. In 1896, Monsieur Baire was lecturing on analysis and carelessly gave the following exercise: to find all solutions ...
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Implication of Baire Category Theorem

The statement that the countable intersection of open dense sets is dense is equivalent to the statement that the countable union of nowhere dense sets contains no balls. When proving the ...