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Questions tagged [the-baire-space]

For questions about the Baire space, that is the family $\mathbb{N}^\mathbb{N}$ of all sequences of natural numbers with the product topology. For questions about the class of Baire spaces - spaces in which Baire category theorem holds - use (baire-category) tag.

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Complement of any dense countable subset of reals is homeomorphic to irrationals

I recently stumbled upon this: For any infinite countable subset $A\subseteq\mathbb R$ such that $\overline A=\mathbb R$, the complement $\mathbb R\setminus A$ is homeomorphic to the Baire space. (Or,...
Martin Sleziak's user avatar
1 vote
0 answers
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Variant of Baire theorem

I consider $(X,d)$ a complete metric space. I have this weak form of the Baire theorem : There does not exist nonempty open subset $O$ of $X$ such that $O=\bigcup_{n\geq 0} F_n$ where the $F_n$ are ...
G2MWF's user avatar
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3 votes
1 answer
126 views

Is this set countable for any function?

Working with CH I order ($\prec$) $^\omega\omega$ of order type $\omega_1$. I let $$f_\alpha:=\min_\prec\{f\in\hspace{1mm}^\omega\omega:\neg (f(n) \leq f_\beta(n)),\forall n\in\omega,\beta\in\alpha\}$$...
Estragon's user avatar
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1 answer
212 views

Can you construct a sequence to witness that this set isn't compact? [closed]

I know that if $K$ is a closed subset of $^\omega\omega=:\mathcal{N}$ and there exists an $f\in\mathcal{N}$ such that $K\subseteq \{g\in \mathcal{N}:g\leq f \}$ (where $g\leq f$ is pointwise), then $K$...
Estragon's user avatar
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2 votes
1 answer
111 views

How is the set $C(f)\cap V$ of second category in $V$?

I am reading the paper "P. S. Kenderov, I. S. Kortezov and W. B. Moors, Continuity points of quasi-continuous mappings, Topology Appl. 109 (2001), 321–346." Just before Theorem 2 of the ...
Ghosh Da's user avatar
1 vote
1 answer
124 views

Must a subset of the real line that is comprised entirely of condensation points be a Baire space?

Let $X$ be a subset of the real line in which every point is a condensation point. Is $X$ a Baire space?
Kiran Antony's user avatar
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0 answers
69 views

Is this proof that the irrational number set is a Baire space correct?

I know there's proof that the irrational number set is a Baire space, but when I tried to prove this, I used a different way. However, I feel my proof is not right. Let $I= \mathbb{R} \backslash \...
M_k's user avatar
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2 votes
2 answers
209 views

Nowhere dense subsets of a dense subspace are nowhere dense in the whole space and vice versa

It seems, with the following lemma, the proposition at the bottom easily follows. If $Y\subset X$ dense. Then, for every nonempty $A\subset Y$, $\text{Int}_Y \left(\text{cl}_Y A\right)=Y\cap \text{...
user760's user avatar
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1 answer
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Equivalent definitions of Baire spaces

We say that a metric space $X$ is a Baire space if there is no open set $E$ such that $$E \subseteq \bigcup\limits_{n\geq 1} F_i,$$ in which each $F_i$ is a closed set with empty interior. Suppose ...
José Victor Gomes's user avatar
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1 answer
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Excercise 15 Rudin functional analysis chapter 2

I am self-studying the book function analysis of Rudin. I got stuck on the final passage of the following exercise. Suppose $X$ is an $F-$space (a topological vector space with a topology induced by a ...
Matteo Aldovardi's user avatar
2 votes
1 answer
72 views

Examples of dense and codense $G_\delta$ subsets of $\mathbb{R}^2$ that are not homeomorphic to $\mathcal{N}=\mathbb{N}^\mathbb{N}$

I have been asked to show that if $X$ is a dense $G_\delta$ subset of $\mathbb{R}$ such that $\mathbb{R}\setminus X$ is also dense in $\mathbb{R}$, then $X$ is homeomorphic to $\mathcal{N}=\mathbb{N}^\...
closedrhombus's user avatar
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0 answers
32 views

Is there an indecomposable topological group structure on the Baire Space?

Follow up to this follow up question. Is there an indecomposable group $G$ ie, a group that can't be written in the form $A\times B$ where both $A$ and $B$ are nontrivial groups along with a topology ...
Carla only proves trivial prop's user avatar
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1 answer
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Are all topological group structures on the Baire Space of the form $G^{\mathbb N}$ for $G$ a countable discrete group?

This question was accidentally trivial: For any countably infinite discrete group $G$ we have that $G^{\mathbb N}$ is a topological group structure on the Baire Space, as pointed in Qiaochu Yuan's ...
Carla only proves trivial prop's user avatar
0 votes
1 answer
70 views

Is there a topological group structure on the Baire Space?

The sum of two irrationals might not be irrational so we can't use that; Also $\mathbb N$ is not a group so there's no obvious way to define a group structure in it seen as $\mathbb N^{\mathbb N}$ ...
Carla only proves trivial prop's user avatar
21 votes
3 answers
956 views

Does there exist a bijective, continuous map from the irrationals onto the reals?

Let $\mathbb{P}$ be the irrational numbers as a subspace of the real numbers. $\mathbb{P}$ is homeomorphic to $\mathbb{N}^\mathbb{N}$, which is also called the Baire space. It is well known, and ...
Ulli's user avatar
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1 vote
0 answers
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Non empty perfect set and it's cardinality in different spaces

Let $(X, \tau) $ be any topological space. $P\subset X$ is called perfect if $P'=P$ where $P'$ is the set of all limit ponits of $P$. If $(X, \tau) $ is a $T_1$ space, then any open set containing a ...
Ussesjskskns's user avatar
1 vote
0 answers
23 views

Homeomorphism between $\mathbb N^\infty$ and a closed subset $\mathbb M$ of $(\mathbb N^\infty)^\infty$

Trying to figure out a proof of a lemma that I'm reading in Stochastic Relations by Ernst-Erich Doberkat. The Baire space, denoted $\mathbb{N}^\infty$, is the infinite product of the natural numbers. ...
Erik Grnl's user avatar
2 votes
1 answer
161 views

Example of function of Baire 3 and Baire 4

i'm looking for explicit examples of real-valued functions of the Baire third and fourth class, without using Borel-measurability but just using some characterization theorem of the previous classes. (...
Luca Cardarelli's user avatar
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0 answers
105 views

Why Conway 13 base function is Baire two?

I'm writing my thesis about Baire Classes. I want to prove that the 13 base Conway function is in the Baire two class. I need a clear proof of this fact, a proof in which is described how to "...
Luca Cardarelli's user avatar
2 votes
2 answers
130 views

Baire space and increasing union of closed subspaces

Let $X$ be a Baire space. Suppose there is an increasing sequence $C_1\subset C_2\subset \cdots $ of closed subspaces of $X$, whose set-theoretical union is $X$. Since $X$ is Baire, we know that some ...
Ken's user avatar
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0 answers
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Can we classify all topological space $(X, \tau) $ where every second category sets are Residual sets?

$(X, \tau) $ be a topological space. $A\subset X$ is Residual if $X\setminus A$ is of first category. In a Baire space, a Residual set is of second category. $A\subset X$ Residual, then $X\setminus ...
Ussesjskskns's user avatar
2 votes
2 answers
195 views

Baire space is homeomorphic to countably many copies of itself

On wikipedia I found that the Baire space $\mathcal{N}$ is homeomorphic to the product of a countable number of copies of itself, however, I haven't been able to find a proof. The Baire space is ...
Antigone's user avatar
  • 115
2 votes
1 answer
55 views

Is $\mathbb{Q} \times \mathbb{Q}$ a $G_\delta$ set?

I can prove that $\mathbb{Q}$ is not a $G_\delta$ set in $\mathbb{R}$. I was applying the same Baire space argument to show that $\mathbb{Q} \times \mathbb{Q}$ is not a $G_\delta$ set. I was thinking ...
Maths_Coffee's user avatar
2 votes
2 answers
61 views

The Baire space is homogeneous,

it's kwown that the Cantor space $2^{\omega}$ is a homogeneous topological space (because it is a topological group). Does anyone have any idea why $\omega^{\omega}$ is a homogeneous topological space?...
gaam2296's user avatar
0 votes
1 answer
124 views

Every compact subset of Baire Space (ω^ω) has empty interior [duplicate]

Here is the Well known proposition. I assumed that K is the compact subset of ω^ω and it contains an open set Ns, so I thought if K is compact so Ns is because Ns is clopen set, If I can not find a ...
Etü's user avatar
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-2 votes
2 answers
157 views

Every compact subset of Baire space $\omega^\omega$ has empty interior. [duplicate]

Here is the well-known proposition, I assumed that $K$ is compact subset of $\omega^\omega$ and it contains an open Set $N_s$, so $N_s$ is clopen and thus it is compact. If I can show $N_s$ can not be ...
norito's's user avatar
3 votes
0 answers
80 views

$\bigcap_{n\in\mathbb{N}}{F_{n}}$ is dense in $X_{0}$

Let $\{(X_{n},d_{n})\}_{n\in\mathbb{N}}$ be a sequence of complete metric spaces. If $\{f_{n}\colon X_{n}\to X_{n-1}\}_{n\in\mathbb{N}}$ is a sequence of functions continuous such that \begin{equation}...
Darkmaster's user avatar
4 votes
1 answer
222 views

$X$, $Y$ be metric spaces and $f: X \to Y$. If $X$ is Baire and $Y$ is separable then $f$ is continuous in a dense $G_{\delta}$ of $X$

Let $X$, $Y$ be metric spaces and $f: X \to Y$. Suppose $X$ is Baire and $Y$ is separable. If $f ^{− 1} (O)$ is $F_{\sigma}$ for every open $O \subset Y$, show that $f$ is continuous in a dense $G_{\...
user1999's user avatar
  • 504
10 votes
2 answers
262 views

If $f_{n}$ has a dense image, then $\bigcap (f_{1}\circ\cdots\circ f_{n})(X_{n})$ is dense

Let $\{(X_{n}, d_{n})\}_{n\in\mathbb{N}}$ be a sequence of complete metric spaces and $\{f_{n}: X_{n}\to X_{n − 1}\}_{n\in \mathbb{N}}$ a sequence of continuous functions. If $f_{n}$ has a dense image ...
User1997's user avatar
  • 391
2 votes
1 answer
40 views

Description of a set in the Baire space.

I have the following exercise. In the Baire space, $\omega ^ \omega$ ,describe a set $A$ such that $int(A)\neq \emptyset$ and $A\neq \bar{A}$ I recently just got to know the Baire space, it is ...
Haus's user avatar
  • 744
0 votes
1 answer
143 views

Difficulty showing a dense $G_{\delta}$ subset of a Baire space is Baire

I have been working on showing that the irrationals is a Baire space. So far I have shown that the irrationals can be expressed as a $G_{\delta}$ set and I know that if this set was to be Baire then ...
CBBAM's user avatar
  • 6,265
0 votes
2 answers
420 views

A $G_{\delta}$ subset of a Baire space is Baire

I have been seeing this fact used a lot but have not been able to find a proper proof justifying it. Would anyone be able to outline one?
CBBAM's user avatar
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3 votes
1 answer
328 views

On Čech-complete space.

I'm reading an article of topology and i came across a Properties : Properties : Closed subspaces and arbitrary products of Čech-complete spaces are Čech-complete Every Čech-complete space is a ...
User77's user avatar
  • 89
1 vote
1 answer
139 views

Recursive representation of closed sets in Baire Space ($\Sigma^1_1$)

These notes (pg 2) say that A set $C \subseteq \omega^\omega$ is closed if and only if there is an $z \in \omega^\omega$ and a recursive predicate $R \subseteq \omega^{<\omega}\times\omega^{<\...
Analytic_Suslin's user avatar
0 votes
1 answer
243 views

Equivalent formulations of analytic/ $\omega$-suslin sets (in Baire Space) [duplicate]

I'm aware that there are more general formulations of these concepts, but I'm just starting to learn these and have been looking at them in a very restricted and simple context. Let $A \subset \omega^{...
Analytic_Suslin's user avatar
0 votes
1 answer
155 views

Does there exist a second category set that is not a Baire space?

Can someone give an example of a second-category set $Y$ in a metric space $X$ but $Y$ is not a Baire space. We know that Baire space$\implies $ Second Category. But I am trying to show that the ...
Kishalay Sarkar's user avatar
0 votes
3 answers
79 views

Clarification of Baire category theorem: a (counter-?)example

I am trying to understand the statement of Baire's category theorem. Why is $\bigcap_{n \in \mathbb{N}} A_n := \bigcap_{n \in \mathbb{N}} \big[(n, n+\frac{1}{2}) \cap \mathbb{Q}\big] = \emptyset$ ...
Friedrich's user avatar
  • 478
0 votes
1 answer
112 views

Wadge Hierarchy on $\Delta^{0}_{2}$ sets

i'm studying the Wadge hierarchy on Baire space and Cantor space. I'm asking whether or not the $\Delta^0_2$ sets form a unique degree in these spaces and why the $\Sigma^0_2$-true (i.e. not polish) ...
Ajeje's user avatar
  • 101
0 votes
1 answer
158 views

Showing $A\cap B \ne \emptyset$

So I've been trying to prove this problem for the past week but I don't know if the path chosen was prudent. I used Baire's Theorem to arrive at my proof for the problem. May someone kindly provide ...
shiloh.otis's user avatar
1 vote
2 answers
52 views

Operation over the Baire space $\omega^\omega$ that preserves Borel sets

We have the following operation on sequences of natural numbers (elements of $\omega^\omega$): $$\begin{align}*:\omega^\omega\times\omega^\omega &\longrightarrow \omega^\omega\\ (x,y) &\...
Lorenzo's user avatar
  • 2,611
4 votes
1 answer
129 views

Continuous parametrization of continuous functions in the Baire space $\omega^\omega$

It can be shown that in the Baire space there is a bijection $$\begin{align}\mathcal{N} &\longrightarrow \{f \in {}^\mathcal{N}\mathcal{N} \mid f \text{ is Lipschitz}\}\\ x &\longmapsto \ell_x\...
Lorenzo's user avatar
  • 2,611
-1 votes
1 answer
30 views

A subset of the second Baire category on the real line

Why is the subset in $\mathbb{R}$ of the second Baire category uncountable?
iamilyaboldyrev's user avatar
2 votes
1 answer
67 views

A question regarding the convergence of a sequence of functions at some point on $[0,1]$

Problem: Does there exist a sequence of continuous functions $ f_n:[0,1] \to [0,\infty)$ such that $\lim_{n \to \infty} \int_0^1 f_n(x) dx=0$ but their doesn't exist any $x \in [0,1]$ for which the ...
Rabi Kumar Chakraborty's user avatar
1 vote
0 answers
217 views

$\Bbb R^J$ is a Baire space in product topology

I have written the following proof of the fact that, $\Bbb R^J$ is Baire space in product topology. Can anyone check my proof and say about any fault? Thanks in advance. $\textbf{Proof :---}$ ...
Sumanta's user avatar
  • 9,634
0 votes
2 answers
2k views

Q(set of rationals) is of baire's first category in itself but N(natural numbers) are of second category in itself .

since we need to see whether they can be written as a countable union of nowhere dense sets or not . for N , i thought {1} these single-tons are dense in N . so N is of second category. Is this ...
Yash Sarin's user avatar
0 votes
1 answer
136 views

Baire space has isolated point

Let $X$ be a Hausdorff, Baire space. I want to prove that $X$ has an isolated point. In a Hausdorff space, singletons $\{x\}$ are closed. Now suppose for a contradiction $X$ has no isolated points. ...
wwinters57's user avatar
1 vote
1 answer
22 views

Convergence in the Baire space $\mathcal{N}=\mathbb{N}^{\mathbb{N}}$

$\mathcal{N}:=\mathbb{N}^{\mathbb{N}}$ Let $d(m,n)=\sum\limits_{n=1}^{\infty}\:\frac{1}{2^n}\: d_n(m_n,n_n)$ be the product metric on $\mathcal{N}$, where $d_n$ denotes the discrete metric. I want to ...
user avatar
0 votes
1 answer
185 views

How to show that $\mathbb{N}$ is a Baire space [duplicate]

I read that a topological space $(X,d)$ is a Baire space if for every sequence $\{X_n\}$ of open dense subsets of $X$, the set $\bigcap_{n=1}^{\infty}X_n$ is also dense in $X$. Since every complete ...
Link L's user avatar
  • 735
3 votes
2 answers
381 views

Polish spaces are continuous images of the Baire space

I'm having some troubles understanding the proof of Theorem 7.9 (pag. 39) in Kechris' "Classical Descriptive Set Theory": There are two points of the proof proposed that I don't quite understand. ...
Lorenzo's user avatar
  • 2,611
0 votes
1 answer
882 views

Let $X$ be a complete metric space. $E$ is a non empty open set in $X$ Then is $E$ a first category set or second category set.

Let $X$ be a complete metric space. $E$ is a non-empty open subset of $X$ Then is $E$ a first category set or a second category set. it seems to me that $E$ should be a second category set, but how ...
Eklavya's user avatar
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