Unions of sets exist? [closed]

Hello,

Probably this questions is very stupid, but anyway: It usually said that the category of sets is cocomplete, in particular meaning that we have disjoint unions of arbitrary families of sets, indexed by a set. I do not have very good intuition about set theory, so the question is:

If we have a family of sets, and somehow the cardinalities of the sets in the family grow very very fast; Could that be that the disjoint union is actually not a set?

Furthermore, it is reasonable that even if the answer is yes, families which are constructed naturally do have some boundness condition which implies that the disjoint union is a set.

Finally, is the use of Grothendieck universes solve this nuisance?

Thank you very much, Sasha

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closed as too localized by Martin Brandenburg, Andrés E. Caicedo, Andreas Thom, Pete L. Clark, Loop SpaceNov 3 '10 at 10:09

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It's an axiom of ZF set theory that the union of a set of sets is a set. Does this help? – Kevin Buzzard Nov 2 '10 at 19:44
The boundedness condition is just that the index class of the family is actually a set. – Stefan Geschke Nov 2 '10 at 21:10
As @Stefan said, the trick is that by "family" we in particular mean set of sets. The idea is that "set"s are somehow guaranteed to be "small", as compared with "classes" or "universes" or whatever language you prefer. A good analogy is with numbers: there are infinite numbers, but any sum of finitely many finite numbers is guaranteed to be finite. Similarly, any union of "setly many" sets is a set. – Theo Johnson-Freyd Nov 3 '10 at 4:07
I voted to reopen. This strikes me as a reasonable question for a mathematician who has a good general background but has never actually learned rigorous set theory. In particular, the discussion about how Replacement fits in is not obvious. I don't know that this question needs any more answers -- Arturo's is very good -- but I don't agree that this sort of question is unwelcome. – David Speyer Nov 3 '10 at 13:06
Thank you all that answered, your answers were very helpful for me, although my struggle with set theory is not over yet :). I do not choose accepted answer, as they all were helpful. – Sasha Nov 3 '10 at 18:19

A family of sets is really a set whose elements are sets. In ZFC, the Axiom of Union states (taken from Jech's Set Theory):

Axiom of Union. For any $X$ there exists a $Y = \bigcup X$.

That is: $$\forall X\\ \exists Y\\ \forall u\\ \left( u\in Y \leftrightarrow \exists z(z\in X\wedge u\in z)\right).$$

So if you have a family of sets, this will play the role of $X$ in the axiom; the sets in the family are the sets $z$. Thus, you cannot have the cardinalities "grow so fast" that the union will not be a set; that can only occur if your collection of sets is not itself a set but a proper class to begin with (e.g., if you tried to take the "union" of the collection of all ordinals), whether the family is disjoint or not.

So now suppose you have a family $X$ of sets, and you want to consider "the" disjoint union of the elements of $X$ (up to a bijection, which are the isomorphisms in the category of sets). Using the Axiom Schema of Replacement, with the function $\mathbf{F}(x) = x\times\{x\}$, there exists a set $Y = \mathbf{F}[X] = \{\mathbf{F}(x)\mid x\in X\}$. Then $Y$ is a set, and the elements of $Y$ are pairwise disjoint: if $\mathbf{F}(x)\cap\mathbf{F}(y)\neq\emptyset$, then there exists $z\in x\times\{x\}$ such that $z\in y\times\{y\}$. So $z=(a,x)$ for some $a\in x$, and $z=(b,y)$ for some $b\in y$; hence $(a,x)=(b,y)$, so $x=y$. So let $Z=\cup Y$. Then $Z$ is a disjoint union of the sets in $X$, and is a set because it is a union of $Y$ (using the axiom of unions), which is itself a set by the Axiom of Replacement.

Edit: Actually, I'm being needlessly complicated in the last paragraph; there is no need to invoke the Axiom of Replacement. Given a family $X$ of sets, we can take $Y=(\cup X)\times X$ (which is a set by the Axioms of Union and Power sets, and the Axiom Schema of Separation) and then take $Z=\{ (a,b)\in Y\mid a\in b\}$. This set achieves the same objective, since for each $x\in X$, the subset $Z_x = \{(a,b)\in Z\mid b=x\}$ is bijectable with $x$, the set $Z$ is the union of the $Z_x$, and $Z_x\cap Z_y\neq \emptyset$ if and only if $x=y$.

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What Grothendieck universes can do for you is that they are sets that are closed under the operation of taking unions of families of sets that are indexed by a set.

In Zermelo-Fraenkel set theory (ZF) the class of all sets is closed under taking unions over families of sets indexed by a set. This follows from the Replacement Scheme (images of sets under functions are sets) and the Axiom of the Sum Set (the union over a set is a set). So Grothedieck universes are sets that act like the class of all sets in ZF.

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I asked a question a few weeks ago that led to the discussion of a similar point: Axiom of Replacement in Category Theory

A set-indexed family is a set because of the axiom of replacement, which is an axiom in standard set theory. So you don't need to bound the size of the sets in your family. Some people don't like the axiom of replacement for the essentially the reason you mention: it allows you to construct sets out of families of sets whose cardinalities grow very fast. An easy example is the family $$\mathbb{N}, \mathbf{P}\mathbb{N}, \mathbf{PP}\mathbb{N}, \ldots$$ where $\mathbf{P}$ is the power-set operation.

If you don't want to use replacement, then you have to explicitly construct a set that contains the sets you want to form. You can do this if you can bound the cardinalities of the sets, so your intuition of the underlying issue is sound. If you can do that, then you only need to invoke the axiom of union. Arturo Magidin gives the details of the construction in his answer.

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It's not clear to me what you are calling a "family of sets". Usually I would think that this is exactly the same as a set of sets. In that case, there is no need for replacement at all. Moreover, the set of cardinalities in any family will be bounded because any set of ordinals has an upper bound. – Carl Mummert Nov 3 '10 at 2:14
@Carl Mummert: The original poster talks about an indexed family of set, which I would interpret as a function $F:I\to V$ for some index class $I$. If $I$ is a set, then $\bigcup_{i\in I}F(i)$ is a set as well, but we need replacement to show this. – Stefan Geschke Nov 3 '10 at 9:39
Just to clarify my previous comment: We need replacement to show that $\bigcup_{i\in I}F(i)$ is a set if $F$ is a function that is a class and $I$ is a subset of its domain. – Stefan Geschke Nov 3 '10 at 9:41