6
$\begingroup$

Reading Halmos' 'Naive Set Theory', I learn that these ZFC set axiom are used to recognise set that are set intuitively.

And the axiom of power, stated as 'For each set $X$, there exists a collection (set) of sets that contains among its elements all the subsets (again a set) of the given set.'

I am actually curious of whether there can be a 'collection' (not a set, it is a 'collection' of elements I have implicitly) 'inside' $X$ (not a subset, since is not a set).

This question may seem strange, one of the motivation is that in the power set $P(X)$, any element are subset of $X$ which is itself a set, I wonder if some 'collection' are already omitted. Another motivation is that any set to be chosen must be 'specific', that is, constucted under a sentence with set-builder notations (under axiom of specification).

I would like to formulate the statement and proof as follows, but fail even to state it.

Statement: Given a 'collection' of element (all $\in X$), it must form a set. (Prove or dispove)

Proof: ...

Problem I find in the statement: I already have in mind a 'collection' of objects, this seems like I can use axiom of specification to construct it as subset. But I really don't know what this set could be, if I really 'know' them, I am actually trying to index them (reducing unknown to known), this is already assuming they form a set (by considering range, which form a set by specification)

The whole problem lies on the definition of 'collection' (not a set). But in ZFC in the book, every objects to study are sets, so it does not really make sense.

Maybe another better way to formulate is if in a larger 'model', can there be defined another other object that is 'inside' $X$ (every element $\in X$) but not a set.

Sorry for being not specific, I will edit it later.

$\endgroup$

3 Answers 3

6
$\begingroup$

Yes, this is possible. One of the ways to see this is the idea of forcing as a tool for adding new sets to the universe.

The first example is adding a new set of integers. As the new set is made only of integers, it is "a collection of integers" but it is not a set in the model. It is not a class either, though. It just exists in a larger universe.

The problem here is that collections might hold information that contradicts with the rest of the axioms of ZF(C). For example, if $M$ is a countable model of ZFC, then there is a bijection between $M$ and the natural numbers (of $M$), which means that there is a collection of integers which codes the whole model.

If we add that collection as a set, then $M$ would be able to recognise itself as a set. Or at least it should. But it really can't. So you will have to violate the axioms of ZFC if you add that as a set.

Finally, it is important to point out that when we have this sort of situation where there is a collection which is a sub-collection of a set, but not a set, then this collection is not definable, and there is no way for the model you're working in to recognise it. In particular, it is not a class either. We are only able to talk about these collections from outside the model.

$\endgroup$
2
  • $\begingroup$ So if such collection is not even definable, it means that the statment is true? $\endgroup$
    – user376921
    Commented Jan 5, 2020 at 13:37
  • 1
    $\begingroup$ It means that this is a complicated issue with subtleties about theory and meta-theory and that you should probably work under the understanding that the answer is "no" until you've learned enough set theory to appreciate why exactly the answer is yes. $\endgroup$
    – Asaf Karagila
    Commented Jan 5, 2020 at 13:39
2
$\begingroup$

You can define, for example, a "class", which is formally nothing more than a predicate, but which we think of "the things which satisfy that predicate". The axiom schema of comprehension tells us that certain classes are in fact sets (namely, if the class is defined as a subclass of a set).

There is a class of all sets: one appropriate predicate is simply "true". The empty set is a class: for example, a predicate would be "false". There is a class of all finite sets: such a predicate would be "there is a bijection from $X$ to some finite ordinal".

$\endgroup$
3
  • $\begingroup$ Maybe the part remains to answer is, can any 'subcollection' in set $X$ be a class (to satisfy certain predicate), hence a set (by axiom of specification)? $\endgroup$
    – user376921
    Commented Jan 5, 2020 at 13:33
  • $\begingroup$ It depends how you're defining what it means to be a subcollection. Any subcollection defined by a predicate is a class - but maybe one can come up with a meaningful kind of "subcollection" that isn't based purely on the ability to recognise the subcollection's elements, and in that case you might not have a class. This is beyond what I've ever thought about, I'm afraid. $\endgroup$ Commented Jan 5, 2020 at 13:36
  • $\begingroup$ Actually I just read Asaf's answer, which I think answers what you wanted. $\endgroup$ Commented Jan 5, 2020 at 13:37
1
$\begingroup$

I am not too familiar with set theory, but I’ll try to answer. In axiomatic set theory, EVERYTHING is a set. Numbers are sets, functions are sets, everything is a set. So talking about a ‘collection’ or ‘element’ doesn’t make sense because they are sets. There are no atoms.

An example of ‘set which is not a set’ (it is called proper class) is the universal class $U=\{x|x=x\}$.

Though it can be written down easily, this thing is not a set. The thing is that you are taking something “too big”, and ZFC axiomatic set theory can’t handle that.

So, to sum up, the very intuitive notion of ‘collection’ is not well defined in axiomatic set theory. Cantor thought something like what you’re thinking, so you’re in good company, but Russel’s paradox broke it down, and indeed in that period started the so called “crisis of the foundations of mathematics”, don’t know if it’s known with this name in English.

If, instead, you start with a set $X$ and take all elements $x\in X$ that satisfy a certain property, then it is a set, but you must restrict to a set $X$ at the beginning to avoid paradoxes. This is actually an axiom of ZFC (maybe the third or fourth, I don’t remeber).

$\endgroup$
3
  • $\begingroup$ I know but is not familiar with proper class, maybe an alternate formulation of my problem is if a set can contains a proper class. $\endgroup$
    – user376921
    Commented Jan 5, 2020 at 13:22
  • $\begingroup$ I don’t think so, but you’d better ask the guys above, they’re clearly more expert $\endgroup$
    – tommy1996q
    Commented Jan 5, 2020 at 13:24
  • $\begingroup$ @LKMKlein: No. It cannot. $\endgroup$
    – Asaf Karagila
    Commented Jan 5, 2020 at 14:44

You must log in to answer this question.