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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Baire related problem

Let $f:\Bbb R\to \Bbb R$, $f$ in $C^{\infty}$. Suppose that for all $x \in \Bbb R$, there exists an integer $n$ (which depends on $x$) such that $f^{n}(x) = 0$ ($f^{n}$ is composing $f$ $n$ times) ...
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Baire space but not locally compact

I need an example that is a Baire space but not locally compact. I think, $\mathbb{R}^ \mathbb{R}$ is such an example. $\mathbb{R}^ \mathbb{R}$ is not locally compact. But I could not proof that it ...
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The union of a sequence of closed sets with empty interiors has empty interior in a compact Hausdorff 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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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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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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1answer
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Category theorem

I don't have a mathematician background (I am engineer) I understand some concepts but still very abstract for me and I have to show the following: 1.- Of what category is the set of all rational ...
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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 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 $G_{\delta}$. ...
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Baire's Theorem proof regarding points revisited

My first question on this point was not answered. Here is the first part of Shilov's proof of Baire's theorem (not an exact lift from the book as I avoided mathematical symbols). I am trying to be ...
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1answer
204 views

Is this Baire theorem a special case of Baire category theorem?

I'm reading a chinese text book "Real Analysis" (by Zhou Minqiang), one of it's conclusion is "Baire theorem" For any $E\subset R^n$ is a $F_\sigma$ set: $$E=\bigcup_{k=1}^{\infty}F_k,$$ where ...
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0answers
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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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Questions related to intersections of open sets and Baire spaces

EDIT: I have reposted this question on MathOverflow. (The version posted there is more concise, with some details omitted. I have also added a question about pseudobases with similar property.) MO ...
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For every $x \in [a,b] , \exists n_x\in \mathbb Z^+$ such that $f^{(n_x) }(x)=0$ ; then to prove $f$ is a polynomial in $[a,b]$

Let $f:[a,b] \to \mathbb R$ be a continuous function having derivatives of all order such that for every $x \in [a,b] , \exists n_x\in \mathbb Z^+$ such that $f^{(n_x) }(x)=0$ ; then how do I show ...
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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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Using Baire Category Theorem to prove $\mathbb{R}^2\not\cong\mathbb{R}^3$.

How can we prove $\mathbb R ^ 2$ is not homeomorphic to $\mathbb R ^3$ using Baire Category Theorem? Here is a standard proof of this fact using algebraic topology. Note that $\mathbb{R}^{3}-\{x\}$ ...
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If every point $x \in X$ has a neighborhood that is Baire space, then $X$ is a Baire space

Show that if every point $x \in X$ has a neighborhood that is Baire space, then $X$ is a Baire space. (Munkres "Topology", 48.3) Here is what I tried : Let $\{U_n\}_{n \geq 1}$ be a collection of ...
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Analogy for Baire categories?

I'm looking for an analogy to grasp the intuitive notion of size that Baire categories on $\mathbb{R}$ provides. For instance, the cardinality of a subset of $\mathbb{R}$ provides a notion of size in ...
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0answers
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Baire Category Theorem with $G_{\delta}$

If we have a subset $Y$ of a complete metric space $X$ such that $Y$ is a $G_{\delta}$, then how would I go to show that $Y$ satisfies the Baire category theorem? I am trying to get there by showing ...
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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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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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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 ...
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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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Showing an operator kernel is nowhere dense…

We are doing a problem that requires us to, for fixed NONZERO element $h \in L^2[0,1]$ (as in, the function $h$ is not zero on a set of positive measure in $[0,1]$) and fixed $f \in L^2[0,1]$ with ...
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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. $E_x$ ...