10
votes
1answer
107 views

Partitions of the Cantor space into parities

Call a partition of $2^\mathbb{N} = A\cup B$ a parity partition if, for any $n\in\mathbb{N}$, flipping the $n$th bit of any element of $A$ results in an element of $B$, and vice-versa. Given a choice ...
5
votes
1answer
88 views

Cardinality of all $\mathbf{\Sigma}^0_\alpha$-sets over Baire space without full choice

It is well-known that the set of all open (or closed) sets on Baire space has cardinality of the continuum. In context of choice, we can prove that the set of all $\mathbf{\Sigma}^0_\alpha$-sets over ...
12
votes
3answers
229 views

Axiom of Choice and Determinacy

In my set theory course we have been talking about the axiom of determinacy. One of the first things we showed was that $AD$ and $AC$ are incompatible. We later showed that $ZF+AD$ implies the ...
4
votes
1answer
135 views

$\bf AD$ implies $\bf AC_{\omega}(\mathbb{R})$?

How to show that $\bf AD$ implies $\bf AC_{\omega}(\mathbb{R})$? $\bf AD$ is abbreviated for axiom of determinacy. $\bf AC_{\omega}(\mathbb{R})$ states that for each family $(X_i)_{i∈\omega}$, in ...
7
votes
1answer
158 views

Constructing a choice function in a complete & separable metric space

Let $X$ be a complete & separable metric space. Let $\{E_i\}_{i\in I}$ be a collection of closed and nonempty sets in $X$. If $X$ is just a complete metric space, it seems not possible to ...
3
votes
1answer
218 views

Lebesgue theory and axiom of choice

I have been told that the existence of non-Lebesgue-measurable sets on $\mathbb R$ is impossible without axiom of choice. Do any other well-known results in Lebesgue theory depend on the axiom of ...
3
votes
2answers
244 views

Proof of Incompatibility of Axioms of Determinacy and Choice

I'm working through some lecture notes on the axiom of determinacy, and have run into some trouble with the proof of the incompatibility of the axiom of determinacy with the axiom of choice. ...
10
votes
1answer
678 views

Constructing a subset not in $\mathcal{B}(\mathbb{R})$ explicitly

While reading David Williams's "Probability with Martingales", the following statement caught my fancy: Every subset of $\mathbb{R}$ which we meet in everyday use is an element of Borel ...
2
votes
2answers
244 views

Dense subset of the plane that intersects every rational line at precisely one point?

It seems there should exist a non-measurable bijection $f: \mathbb{R}\rightarrow \mathbb{R}$. And thus we can obtain a non-measurable graph on $\mathbb{R}^2$ which intersects every horizontal or ...