# Tag Info

17

It is not strictly true from a rigid formal viewpoint. Consider the following property: $$\psi(x) \equiv (\mathrm{AC} \land x\text{ is a Vitali set}) \lor (\neg\mathrm{AC} \land x = \mathbb R)$$ where $\rm AC$ is the formal statement of the Axiom of Choice. ZF proves $\exists x.\psi(x)$, so let $A$ be some set such that $\psi(A)$, and define $f$ to be its ...

11

As Henning said, formally speaking, this is not quite true. We can define sets, without using the axiom of choice, which we cannot prove that they are measurable. For example, every universe of set theory has a subuniverse satisfying the axiom of choice, in a very canonical way, called $L$. We can look at a set of reals which is a Vitali set in $L$, or any ...

9

This is a confusing matter, mainly because the kind of reasoning you use in your proof is usually taken to be valid. However, in order to formalize that reasoning in axiomatic set theory, we need to reduce it to particular symbolic formulas in a formal logic system. And it turns out that the rules of symbolic logic and set theory that are sufficient to ...

8

First, consider that constructing a non-measurable function is equivalent to constructing a non-measurable set (using indicator functions). Second, what you are looking for appears to be Solovay's Theorem: http://mathoverflow.net/questions/42215/does-constructing-non-measurable-sets-require-the-axiom-of-choice https://en.wikipedia.org/wiki/Solovay_model ...

5

The mostly commonly used AC has 3 forms: choice function: suppose you have a bag of inhabited sets $A$, then you can just say: oh, let $f$ be a function on $A$, with the action $f(x)\in x$ for each $x\in A$. zorn's lemma: suppose you have a partial order $A$ with the property that every linearly ordered subset (chain) is bounded above, then you can say: ...

4

Yes, this requires the full strength of the axiom of choice. The proof goes through the following theorem. The following are equivalent of $\sf ZF$. The axiom of choice. If $V$ is a vector space, and $W\subseteq V$, then $W$ has a direct complement in $V$. Now it's easy, because given any $W$, consider the obvious map from $V$ to $V/W$, $V\... 4 In$\sf ZF$the two are equivalent: The axiom of choice. Every$\beth$number is an$\aleph$number. You can prove this by going through the following equivalent statement, The power set of an ordinal can be well-ordered. So if the axiom of choice fails, we know that there is some$\alpha$such that$\beth_\alpha$cannot be well-ordered. ... 3 The issue is that without Dependent Choice stating that every chain is finite does not imply the existence of a maximal element. It is a weird situation, because it means that every finite chain can be extended. However if you have an infinite sequence given, then you do know it is eventually stable. The proof relies on being able to choose some arbitrary ... 2 The whole point of "indexed family" is that it is a function. It is a function mapping the index to some object. So if you consider the first statement, it just says in a more complicated language that every family of non-empty sets has a choice function. Why more complicated? Because it requires the understanding that an indexed family is a function, ... 2 Well you need axiom of choice to construct a non measurable set (see http://mathoverflow.net/questions/42215/does-constructing-non-measurable-sets-require-the-axiom-of-choice). If you construct a function without the axiom of choice then it must be measurable because otherwise we could then construct a non measurable set (as the inverse image of some ... 2 For every$\alpha$there is a proper class of subsets of$X$which can be mapped onto$V_\alpha$, and for each of those there are many surjections. You need to choose a set for each$\alpha$simultaneously and choose a surjection, and do it in a coherent way. Not to mention, that in order to prove that there is a bijection between$X$and$V$you need to ... 1 You’re given a filter$\mathscr{F}$on$X$, and you want to show that there’s an ultrafilter$\mathscr{U}$on$X$such that$\mathscr{F}\subseteq\mathscr{U}$. Since you’re going to be using Zorn’s lemma, it’s certainly reasonable to suppose that$\mathscr{U}\$ will be a maximal element some partial order, but you need to make sure that this maximal element ...

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