All Questions
21
questions
3
votes
1
answer
114
views
Does there exist $f:\Bbb{R}\to \Bbb{R}$ additive onto function such that $f(F) \subset \Bbb{R}$ has the property of Baire for every $F$?
Let $F\subset \Bbb{R}$ intersect every closed uncountable subsets of $\Bbb{R}$.
Does there exist $f:\Bbb{R}\to \Bbb{R}$ additive onto function such that $f(F) \subset \Bbb{R}$ has the property of ...
- 317
1
vote
0
answers
154
views
Study of the class of functions satisfying null-IVP
$\mathcal{N}_u$ : Class of all uncountable Lebesgue-null set i.e all uncountable sets having Lebesgue outer measure $0$.
Let $f:\Bbb{R}\to \Bbb{R}$ be a function with the following property :
$\...
- 317
2
votes
1
answer
133
views
Borel $\sigma$-algebras on paths of bounded variation
Let $(C, \|\cdot\|)$ be the Banach space of continuous paths $x: [0,1]\rightarrow\mathbb{R}^d$ starting at zero with sup-norm $\|\cdot\|$.
Let further $B\subset C$ be the subspace of $0$-started ...
- 411
1
vote
1
answer
161
views
Topological analog of the Lusin-N property
$A\subset \Bbb{R}$ is meager if $A$ can be expressed as a countable union of nowhere dense sets.
Let $f:[a, b]\to \Bbb{R}$ is absolutely continuous, i.e., for every $\epsilon>0$, there exists $\...
- 317
2
votes
0
answers
62
views
Separately continuous functions of the first Baire class
Problem. Let $X,Y$ be (completely regular) topological spaces such that every separately continuous functions $f:X^2\to\mathbb R$ and $g:Y^2\to \mathbb R$ are of the first Baire class. Is every ...
- 4,395
6
votes
1
answer
378
views
What is the Borel complexity of this set?
Problem. What is the Borel complexity of the set
$$c(\mathbb Q)=\{(x_n)_{n\in\omega}\in\mathbb R^\omega:\exists\lim_{n\to\infty}x_n\in\mathbb Q\}$$
in the countable product of lines $\mathbb R^\omega$?...
- 40.2k
1
vote
1
answer
219
views
Is there a simple proof that proves $C^1[0, 1]$ is $\Sigma^1_1$ in $C[0, 1]$?
In his book, "Descriptive Set Theory", Moschovakis states $C^1[0, 1]$ is $\boldsymbol{\Sigma}^1_1$ in $C[0, 1]$ in the exercise 1E.8.
Here, $C[0, 1]$ is the space (metrized by the sup norm) of ...
- 111
9
votes
2
answers
459
views
Small uncountable cardinals related to $\sigma$-continuity
A function $f:X\to Y$ is defined to be
$\sigma$-continuous (resp. $\bar \sigma$-continuous) if there exists a countable (closed) cover $\mathcal C$ of $X$ such that the restriction $f{\restriction}C$ ...
- 40.2k
0
votes
1
answer
147
views
Reference request: Baire class 2 functions
There are many articles on Baire 1 functions, but not many on Baire 2 and above. Where can I find a nice comprehensive survey of them?
- 2,029
19
votes
1
answer
545
views
Can an injective $f: \Bbb{R}^m \to \Bbb{R}^n$ have a closed graph for $m>n$?
Question. Suppose $m>n$ are positive integers. Is there a one-to-one $f: \Bbb{R}^m \to \Bbb{R}^n$ such that the graph $\Gamma_f$ of $f$ is closed in $\Bbb{R}^{m+n}$?
Remark 1. The answer to the ...
- 16.5k
3
votes
1
answer
267
views
Example of a Baire Class $1$ function $f$ satisfying $\omega\cdot n<\beta(f)\leq \omega\cdot (n+1)$ for some natural number $n\geq 1.$
Definitions: Let $X$ be a Polish space (separable completely metrizable topological space).
A function $f:X\to\mathbb{R}$ is Baire Class $1$ if it is a pointwlise limit of a sequence of continuous ...
- 603
9
votes
1
answer
642
views
Examples of Baire Class $\xi+1$ but not $\xi$ functions for each countable ordinal $\xi.$
We say that $f:\mathbb{R}\to\mathbb{R}$ is of Baire Class $1$ if it is a pointwise limit of a sequence of continuous functions.
One can generalize the definition above by taking pointwise limit of ...
- 603
5
votes
1
answer
365
views
Equivalent of Lusin's Theorem in Borel setting
Let $X$ be a Polish space, $\mathcal B$ the sigma-algebra
of Borel sets. Let $E$ be an
aperiodic countable Borel equivalence relation on
$X \times X$ (this means that every class of equivalence
...
- 53
13
votes
0
answers
382
views
Is there a continuous map $f:\mathbb R^\omega\to\mathbb R^\omega$ with dense countable preimage $f^{-1}(\mathbb Q^\omega)$?
Let $\mathbb Q^\omega_0:=\{(x_i)_{i\in\omega}\in\mathbb Q^\omega:\exists n\in\omega\;\forall m\ge n\;\;x_m=0\}$ and observe that $\mathbb Q^\omega_0$ is a countable dense set in $\mathbb R^\omega$ (...
- 40.2k
17
votes
1
answer
756
views
Is there a continuous function $f:\mathbb R^\omega\to\mathbb R$ with injective restriction $f|\mathbb Q^\omega$?
Question. Is there a continuous function $f:\mathbb R^\omega\to\mathbb R$ whose restriction $f|\mathbb Q^\omega$ is injective?
- 40.2k
8
votes
1
answer
363
views
Is each $G_\delta$-measurable map $\sigma$-continuous?
Definition. A function $f:X\to Y$ between topological spaces is called
$\bullet$ $G_\delta$-measurable if for each open set $U\subset Y$ the preimage $f^{-1}(U)$ is of type $G_\delta$ in $X$;
$\...
- 40.2k
1
vote
1
answer
233
views
Definition of $F_{\sigma}$ sets in terms of $\varepsilon$?
Let $X$ be a metric space.
In Borel hierarchy, $\Sigma_{1}^0$ is the set of all open sets in $X$ while $\Pi_{1}^0$ is the set of all closed sets in $X.$ Then at next level, one has $\Sigma_{2}^0 = \{...
- 603
5
votes
1
answer
441
views
Covering measure one sets by closed null sets
(The following question arose in a joint research with Adam Przeździecki and Boaz Tsaban.)
For a $\sigma$-ideal $\mathcal{I}$ of subsets of the unit interval
$[0,1]$, define
$$\newcommand{\card}[1]{\...
- 145
14
votes
3
answers
799
views
Is there a Borel subset of $ \mathbb{R}^{2} $, with finite vertical cross-sections, whose projection onto the first component is non-Borel?
This question is related to another one that I asked two days ago.
Question. Does there exist a Borel subset $ M $ of $ \mathbb{R}^{2} $ with
the following two properties?
The ...
- 1,386
6
votes
1
answer
175
views
On continuous perturbations of functions of the first Baire class on the Cantor set
Is it true that for any function of the first Baire class $f:X\to\mathbb R$ on the Cantor cube $X=2^\omega$ there is a continuous function $g:X\to[0,1]$ such that the image $(f+g)(X)$ is disjoint with ...
- 40.2k
12
votes
2
answers
583
views
Partition $\Bbb{R}$ into a family of sets each one homeomorphic to the Cantor set
It is known that there is no (nontrivial) partition of $\Bbb{R}$ into a countable number of closed set. But is there a partition of $\Bbb{R}$ into sets, each one homeomorphic to the cantor ternary set?...
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