Unions of sets exist? - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-20T07:09:22Z http://mathoverflow.net/feeds/question/44605 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/44605/unions-of-sets-exist Unions of sets exist? Sasha 2010-11-02T19:37:08Z 2010-11-03T06:27:18Z <p>Hello,</p> <p>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:</p> <p>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?</p> <p>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.</p> <p>Finally, is the use of Grothendieck universes solve this nuisance? </p> <p>Thank you very much, Sasha</p> http://mathoverflow.net/questions/44605/unions-of-sets-exist/44610#44610 Answer by Stefan Geschke for Unions of sets exist? Stefan Geschke 2010-11-02T20:11:20Z 2010-11-02T20:11:20Z <p>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. </p> <p>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.</p> http://mathoverflow.net/questions/44605/unions-of-sets-exist/44611#44611 Answer by Arturo Magidin for Unions of sets exist? Arturo Magidin 2010-11-02T20:12:28Z 2010-11-03T06:27:18Z <p>A family of sets is really a set whose elements are sets. In ZFC, the Axiom of Union states (taken from Jech's <strong>Set Theory</strong>):</p> <blockquote> <p>Axiom of Union. For any $X$ there exists a $Y = \bigcup X$.</p> </blockquote> <p>That is: $$\forall X\ \exists Y\ \forall u\ \left( u\in Y \leftrightarrow \exists z(z\in X\wedge u\in z)\right).$$</p> <p>So if you have a <em>family</em> 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.</p> <p>So now suppose you have a family $X$ of sets, and you want to consider "the" <em>disjoint</em> 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.</p> <p><strong>Edit:</strong> 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 $x=y$. </p> http://mathoverflow.net/questions/44605/unions-of-sets-exist/44613#44613 Answer by arsmath for Unions of sets exist? arsmath 2010-11-02T20:49:26Z 2010-11-02T20:49:26Z <p>I asked a question a few weeks ago that led to the discussion of a similar point: <a href="http://mathoverflow.net/questions/41118/axiom-of-replacement-in-category-theory" rel="nofollow">http://mathoverflow.net/questions/41118/axiom-of-replacement-in-category-theory</a></p> <p>A set-indexed family is a set because of the <a href="http://en.wikipedia.org/wiki/Axiom_schema_of_replacement" rel="nofollow">axiom of replacement</a>, 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.</p> <p>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 <a href="http://en.wikipedia.org/wiki/Axiom_of_union" rel="nofollow">axiom of union</a>. Arturo Magidin gives the details of the construction in his answer.</p>