# Tagged Questions

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### 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 ...
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### 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 ...
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### 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 ...
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### $\bf AD$ 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 ...
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### 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 ...
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### 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 ...
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### 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. ...
### 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 ...
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 ...