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.

learn more… | top users | synonyms

2
votes
1answer
27 views

Infinite intersection between a arbitrary set of integers and a set of floor powers

Let $E$ be an infinite set of positive integers, proves that there is a $\alpha \in \mathbb{R}$ such that $\{\left \lfloor \alpha^k \right \rfloor ;k \in \mathbb{N} \}\cap E$ is infinite. I have two ...
1
vote
1answer
36 views

Equivalence of Forms of Baire Category Theorem

I am trying to show the equivalence of two forms of the Baire Category Theorem. These are the two statements: Let $(X,d)$ be a complete metric space. Let $U_n$ be a dense, open set for each $n \in ...
4
votes
0answers
48 views

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 ...
1
vote
1answer
30 views

Convergence of a sequence holomorphic functions

Let $f_n\in\mathcal{O}(\Omega)$ be a sequence of holomorphic functions, s. th. $f_n\rightarrow f$ pointwise in $\Omega$. Show that exist open and dense set $\Omega'\subset \Omega$ such that $f_n$ is ...
4
votes
1answer
100 views

Showing the sum of two sets contains an interval - Baire's Theorem?

If $E$ and $F$ are measurable subsets of $\mathbb{R}$ and $m(E), m(F) >0,$ then $E+F$ contains an interval. The path to the standard solution to this is built on the notion that a measurable ...
0
votes
1answer
53 views

Open mapping lemma - are these versions equivalent?

Here is a version the Open Mapping Lemma given in class : Let $X$ be a Banach space and $Y$ be a normed space. Let $T : X\rightarrow Y$ be a bounded linear map. Assume there exist $M \geq 0$ and ...
2
votes
2answers
55 views

Strong version of Baire Category Theorem

We know that in a complete metric space or compact Hausdorff space the intersection of $\omega$-many open dense sets is dense. In such spaces is the intersection of fewer than $2^\omega$-many open ...
4
votes
2answers
86 views

Baire's property iff first category has dense complement.

Show that $(S, d)$ has Baire's property iff every set of first category has a dense complement. A set is of first category if it is a countable union of nowhere dense sets. First Category Baire's ...
0
votes
1answer
23 views

Two proofs with possibly Baire category theorem about completness.

I'm working with completness right now and I've come across two interesting problems. In my opinion they are worth a little bit attention . a) Let $K$ be closed subset with empty interior on ...
0
votes
1answer
23 views

Baire category theorem in use on a plane

Let $F\subset\mathbb{R}$ be a closed nowhere dense set. One must show there exists $(a,b)\in S^1$ for which $b\neq qa+c$, for all $q\in\mathbb{Q},c\in F$. It's my second question concerning Baire ...
5
votes
3answers
133 views

baire category and the union of dyadic balls of rational center

Suppose that $\{r_n\}_{n=0}^\infty$ is an enumeration of $\mathbb{Q}^N$ and $U = \bigcup_{n=0}^\infty B(r_n,2^{-n})$. We can use a trivial measure theory argument to prove that $U \neq \mathbb{R}^N$. ...
0
votes
0answers
27 views

Continuity Points of The Deriviative of a Differentiable Function

I recently read the section on Baire Spaces from Munkres and this question came to my mind which I cannot settle. Does there exist a differentiable function $f:\mathbf R\to\mathbf R$ such that the ...
3
votes
1answer
30 views

For $f :\mathbb R \to \mathbb R $, there exists an $(a,b)$, such that $f$ is bounded on a sequence with limit $x$, for all $x\in(a,b)$

I want to prove the following. Let $f : \mathbb R \mapsto \mathbb R $. Show that there exists an interval $(a,b) \in \mathbb R $ and $c >0 $: such that for any $x \in (a,b) $ there is a sequence ...
2
votes
3answers
114 views

Proof that every closed subset of $\mathbb R$ is finite or countable or continuum.

I want to prove that every closed subset of $\mathbb R$ is finite or countable or continuum. I know that for arbitrary subset we can not make similar statements - because of continuum hypothesis. ...
0
votes
1answer
27 views

how to show boundary of either open or closed set is nowhere dense.

how to show boundary of either open or closed set is nowhere dense. i think we need to use baire category thm? countable intersection of dense, closed set is once again a dense, closed (and ...
1
vote
1answer
31 views

If $\mathbb{R}=\bigcup_{n=1}^{\infty}E_n$, then the closure of some $E_n$ contains an interval

I'm working on some problems in Carothers' Real Analysis. I just started the section on the Baire Category Theorem. Thus far Carothers has given the Baire Category Theorem for $\mathbb{R}$. Prove ...
1
vote
0answers
215 views

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 $\mathbb{R}^{3}-\{x\}$ ...
0
votes
2answers
27 views

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) ...
2
votes
1answer
34 views

Non-decidable $\Pi^0_1$ (effectively closed) classes

Are there non-decidable $\Pi^0_1$ (effectively closed) classes? According to a draft of Effectively closed sets by Cenzer and Remmel, the class $$ P = \{ 0^n1^\omega \mid n \in B\} $$ is a ...
1
vote
1answer
46 views

Is the union of all $l^p$ spaces meagre in $l^\infty$?

Is the union of all $l^p$ spaces meagre in $l^\infty$? i.e. is $ \bigcup_{p=1}^\infty l^p$ meagre? I am revisiting this variety of math after a long break so help is appreciated. Please correct ...
1
vote
2answers
51 views

Is $c_0$ meagre in $l^\infty$?

I have been reading up on the meagre sets sans an instructor after a 10 year break from this sort of mathematics and I would like to test my understanding. I want to know if the following is true (and ...
3
votes
2answers
92 views

Showing a set is nowhere dense in $C([0,1])$

Let $E_n$ be the set of all $f \in C\big([0,1]\big)$ for which there exists $x_0 \in [0,1]$ (depending on $f$) such that \begin{align*} \lvert\, f(x)-f(x_0)\rvert \leq n\lvert x-x_0\rvert, ...
0
votes
0answers
65 views

A Property of Baire Spaces

Let $X$ be a topological space. I define $X$ to have Property A provided that every closed meager subset of $X$ is nowhere dense. It is easy to see that all Baire spaces have Property A. Is the ...
2
votes
1answer
175 views

In a complete metric space with no isolated points, any countable intersection of open dense sets is uncountable?

I was playing with Baire's Theorem, and seemed to deduce the following: In a complete metric space $X$ that has no isolated points, any countable intersection of open dense sets is uncountable. ...
0
votes
1answer
46 views

A homeomorphism of a topological space with itself maps a set into one of the same category

Prove: If $h$ is a homeomorphism of $S$ onto $S$ and if $E\subset S$, then $E$ and $h(E)$ have the same category in $S$. Rudin, Functional Analysis, 2/e, p.43. (My own answer follows below.)
2
votes
1answer
32 views

Baire Category for monotonic sequence

Let X be a non-empty complete metric space and let $\{f_n:X\to \Bbb R\}^\infty_{n=1}$ be a sequence of continuous functions with the following property: for each $c\in X$, there exists an integer ...
1
vote
2answers
38 views

Is the complement of a set 1st category set $X$ of 2nd category?

If I have some metric space $M$ and find that $X \subset M$ is of 1st category (resp. 2nd category), is the complement of $X$, $X^c$ of second cateogry (resp. 1st category)? Thanks!
0
votes
1answer
29 views

How can the Baire Category Theorem be used to show a point in a complete metric space has some particular property?

My book gives a very tantalizing yet brief note that the Baire Category Theorem can be used to show that a point $x$ in a complete metric space $M$ has a particular property $P$. It states: "A ...
0
votes
1answer
35 views

A set that is a countable intersection of open and dense sets but not open.

We know that Baire Category Theorem implies that in a complete metric space, the countable intersection of open and dense sets is nonempty and actually dense itself. But it is clear that a countable ...
1
vote
1answer
52 views

Why is the subspace of polynomials with degree $\leq$ n nowhere dense in $\mathbb{R}[X]$?

There's a popular application of Baire's Category Theorem that shows that $\mathbb{R}[X]$ (the space of all polynomials with real coefficients) is not complete by showing it is a countable union of ...
1
vote
2answers
71 views

Baire's Theorem Formulation

I am still having problems with some of the proofs of Baire's theorem. In Introductory Real Analysis by Kolmogorov and Fomin The statement of the theorem states that: "A complete metric space R ...
-1
votes
1answer
37 views

Is complement of a 1st category set in a 2nd category space, dense?

Let $X$ be a second category space and $Y$ be a first category subspace of $X$. Is $X \setminus Y$ dense in $X$?
1
vote
1answer
49 views

B meager, not empty, not open implies complement is dense

Let X be topological space. The subset B is meager, not open and non-empty. How can you prove that the complement is dense?
0
votes
1answer
48 views

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 ...
0
votes
1answer
47 views

Baire category theorem to show a set is dense.

Consider $A_j$ a sequence of subsets of $[0,1]$ s.t. for each $N\geq 1$, $\bigcup_{j=N}^\infty A_j$ is open and dense in $[0,1]$. If $S$ is the set of points $x \in [0,1]$ s.t. $x \in A_j$ for ...
1
vote
0answers
51 views

how do i prove that the the set of irrationals cannot be a countable union of closed subsets? [duplicate]

Let $\mathbb{R}$ be equipped with the standard topology. Let $E$ be the set of irrational numbers. How do i prove that $E$ is not a countable union of closed subsets, using Baire Category Theorem?
1
vote
0answers
40 views

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 ...
4
votes
1answer
115 views

Nowhere dense set…

Let $A_n$ be a subset of continuous functions on $[0,1]$ given by: $A_n$ = {$f∈C[0,1]$:there exists $x∈[0,1]$ such that $|f(x)−f(y)|≤n|x−y|$ for all $y∈[0,1]$}. Show $A_n$ is nowhere dense, and use ...
0
votes
0answers
24 views

Uniform boundedness… [duplicate]

Let $\{f_n\}$ be sequences of real valued continuous functions on $\Bbb R$ that is convergent pointwise on $\Bbb R$. Show that there exists $M>0$ and interval $I\subset \Bbb R$ such that ...
3
votes
1answer
100 views

A Baire category question

Let ${f_n}$ be a sequence of real valued continuous functions converging pointwise on $\Bbb R$. Show that there exists a number $M>0$ and an interval $I \subset \Bbb R$ such that $\sup\{ |f_n(x)|:x ...
7
votes
0answers
353 views

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 ...
2
votes
0answers
38 views

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 ...
3
votes
1answer
53 views

Linear functional in Banach space

Let $X$ be a Banach space, $(f_n)\in X^{*}$ a sequence with $f_n\neq 0 $ $ \forall n\in \Bbb N$. Show that there is a $x\in X$ such that $f_n(x)\neq 0 $ $\forall n$. Need some help. Thank you!
1
vote
1answer
118 views

corollary to baire category theorem

I'm studying topology with gamelin and greene's text and I came across a corollary to the baire category theorem which states that "Let (En) be a sequence of nowhere dense subsets of a complete ...
2
votes
0answers
29 views

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 ...
2
votes
1answer
66 views

Equivalence of Baire Space definitions

I am hoping someone could help me show that the following statements, which define a Baire Space, are equivalent. Defn1: Any topological space X such that the intersection of any countable ...
0
votes
0answers
79 views

Kuratowski-Ulam Theorem

I am looking for a proof of the Kuratowski-Ulam Theorem: Let $X,Y$ be Baire spaces, $Y$ be second countable and let $R \subseteq X \times Y$ have the Baire property. Then $\{ x \in X \ | \ R_x ...
0
votes
1answer
117 views

Decomposing real line as a union of a nullset and a set of first category

$\Bbb R$ can be written of the form $A\cup B$ such that $A$ is of measure zero and $B$ is of the first category! can anybody prove this?? I guess $A$ must be an $G_{\delta}$ set which is dense in ...
0
votes
1answer
38 views

Application of Baire's theorem

Let $f: (a,b) \rightarrow \Bbb R$ be a differentiable function in $(a,b)$. Calculate the pointwise limit of: $$f_n(x)=n(f(x+1/n)-f(x)), x\in(a, b-1/n). $$ Let $E_n$ be a countable ...
0
votes
2answers
50 views

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 ...