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41
votes
2answers
3k views

Is every sigma-algebra the Borel algebra of a topology?

This question arises from the excellent question posed on math.SE by Salvo Tringali, namely, Correspondence between Borel algebras and topology. Since the question was not answered there after some ...
2
votes
1answer
317 views

Different Metrics for Baire Space and their induced Topologies

The Baire-Space is the set of all infinite sequences of integers, i.e. $$ \mathcal N = \omega^{\omega}. $$ On this space usually the following metric is given $$ d(\alpha, \beta) = \left\{ ...
26
votes
0answers
806 views

Concerning the various proofs from the axiom of choice that R^3 admits of surprising geometrical decompositions into circles, skew lines and so on: can we prove in any instance that there are no Borel such decompositions? Or that AC is required?

This question follows up on a comment I made on Joseph O'Rourke's recent question, one of several questions here on mathoverflow concerning surprising geometric partitions of space using the axiom of ...
23
votes
4answers
6k views

Non Borel sets without axiom of choice

This is a simple doubt of mine about the basics of measure theory, which should be easy for the logicians to answer. The example I know of non Borel sets would be a Hamel basis, which needs axiom of ...
14
votes
3answers
1k views

Parts of Set Theory immune to independence

The motivation for asking this question is a passage (3.2) in an article by Greg Hjorth where he said that "...it is also an attractive feature of the theory of Borel cardinalities and of the theory ...
10
votes
1answer
380 views

Restrictions of null/meager ideal

Let I denote the null (resp. meager) ideal on reals. Is it consistent that for any pair of non null (resp. meager) sets A and B, there is a null (resp. meager) preserving bijection between A and B? In ...
8
votes
1answer
220 views

Can $\mathbb{R}$ be partitioned into dedekind-finite sets?

Assuming $ZF$ itself is consistent, it is consistent that there are sets $D$ which are infinite but cannot be placed in bijection with any of their proper subsets; such sets are called "strictly ...
4
votes
1answer
157 views

Uncountable atomless subalgebras of the Boolean algebra of all Jordan measurable sets in [0,1]

Definition: Suppose $\mathcal A$ is the Boolean algebra of all Jordan measurable sets in $I=[0,1]$ (i.e $\mathcal A=\{A\subseteq I: \mu(\partial(A))=0\}$, where $\mu$ is the Lebesgue measure and ...
2
votes
0answers
148 views

How many Dedekind-finite sets can $\mathbb{R}$ be partitioned into?

Building off Asaf Karagila's answer to my previous question (Can $\mathbb{R}$ be partitioned into dedekind-finite sets?) on partitioning $\mathbb{R}$ into strictly Dedekind-finite sets: (1) What ...