What does it mean for a set to exist?
I have seen the axiom of choice stated that: for any collection of sets $\{S_\alpha\}_{\alpha\in A}$ there exists a set $B=\{b_\alpha \mid b_\alpha\in S_\alpha \text{ for all } \alpha\in A\}$. So in effect we are "choosing" an element from each $A_\alpha$ and forming the set $B$ out of those elements.
My main question is:
1. What does it mean for this set $B$ to exist? Generally, what does it mean for a set to exist?
Of course this may really be about the concept of existence in mathematics. I don't want to get all metaphysical or philosophical here (which is ironic, because I usually do want to get all nonsensical-philosophical). I'm looking for a direct mathematical answer/explanation.
Here is my intuition: in mathematics, we are always working with a collection (set?) of mathematical objects and their properties/relations (e.g. numbers, or sets of certain symbols, etc.). So for a set to exist, it just means that it is a member of the universe of sets that we are working in (or be in whatever model of whatever set theory we are working with).
E.g. if $\{\emptyset, \{a,b\}\}$ is our universe, then the set $\{a\}$ does not exist.
As a related side question, if we allow the power set of real numbers to exist, does that imply the axiom of choice? It seems like it should because for whatever collection of sets of real numbers we have, any arbitrary set with one element taken form each will be an element in the power set.
These probably have answers already here on MathSE, but this is far out of my expertise, so I have trouble understanding many of the more technical set-theoretic threads. I'd appreciate as simple an answer as possible, but do also appreciate any technical details, even if I don't understand them right away.
Your intuition is generally right: To say that something "exists" in set theory (which is the same as saying that is "is a set") means just that it is an individual in whatever model or interpretation of the language of set theory we're considering.
When we're looking at the model from the outside, it may be possible that we can identify certain of its individuals by a particular property, and then we can ask whether all those individuals make up a set in the model -- which is nothing more or less than asking whether there is some individual in the model whose elements are exactly the ones we were pointing at before. This may or not be the case.
In particular, Russell's paradox points out that if we're looking at all the individuals in the model that are not elements of themselves (for all we know, that could be all of them!), then there is no set in the model that has exactly those individuals (and no others) as elements.
Note well that in this model-theoretic view all the sets in the model have already existed all the time. Thus the axiom of choice does not actively create the choice set -- it just asserts that there is one somewhere in the model. And if it is right about asserting that, then the choice set has always been part of that model.
Right. Though if that is the case, and neither of $a$ or $b$ happens to be $\emptyset$, then it is a strange set-theoretic universe because neither of its two individuals is an element of the other. So as far as the language of set theory is concerned, both $\emptyset$ and $\{a,b\}$ qualify as empty sets in your model, and the Axiom of Extensionality fails for this universe. That is, as long as $\in$ of your model is interpreted to mean the $\in$ relation we use outside it.
No, that is a misunderstanding of what the power set axiom says. What it says is that whenever $X$ is a set in our model, then there is another set $Y$ in the model, such that the elements of $Y$ is exactly all of those individuals in the model that happen to be subsets of $X$.
It may be that from the outside we can point to certain of the elements of $X$, such that there is no set in the model that contains just the elements we point to. That is not a problem for the power set, because the power set is not obliged to contain things unless they are in the model in the first place!
Another way of looking at it is if we stand outside the model and point to certain of the elements of $X$, and if it so happens that there is a $Z$ in the model whose elements are exactly the ones we point to then this $\mathcal P(X)$ has to contain $Z$ as an element. But if the second of these conditions are not met, then the power set axiom demands nothing.