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"A category is said to be small if its objects form a set."

Now one question is in my mind and that is although we know lots of sets and always working with them, but how we can show a class of objects form a set with proof. Some methods are easy and I know, like;

1- If we encounter to Countor contradiction we can say that our class of objects is not a set.

2- If we can earn our class of objects with getting union, intersection and some set operators from another well-known sets, by ZF axioms we can claim our class of objects is a set.

But I'm interest to have a method or definition if exists for use in all cases. If anybody knows more methods or even any other special cases, I will be glad to know.

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There is nothing specific to category theory about this problem. But here are some general guidelines on when a class $A$ is a set in ZF: 1. If $A$ has members of arbitrarily large cardinality, then it is not a set. 2. If $A$ has members of arbitrarily large rank, then it is not a set. etc. –  Zhen Lin Feb 15 '13 at 12:57
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Basically, you're allowed to use unions and powersets to get up to the cardinality you need, starting from any collection that is already known to be a set. Then you make use of the axiom of replacement to show that the collection you're specifically interested in forms a set. –  goblin Feb 15 '13 at 13:07
    
Is there any nonset category with natural objects and structures? –  omid saba Mar 15 '13 at 8:19

1 Answer 1

up vote 5 down vote accepted

The general method to show that a collection $C$ is a set is to show that it is a subset of some set $B$ which you already know is a set.

This corresponds to a rigorous criterion for sethood: a collection $C$ is a set if and only if the rank of its elements is bounded. On one hand, if $C$ is a set, then the rank of $C$ is a bound for the rank of all of its elements. Conversely, if the rank of all the elements of $C$ is bounded by $k$ then $C$ is a subset of $V_{k+1}$, which is a set in the cumulative hierarchy.

In practice, however, it is more common to simply identify an arbitrary set $B$ that has $C$ as a subset, rather than computing a bound for the ranks of elements of $C$. For example, the collection $F$ of all functions from $\mathbb{R}$ to $\mathbb{R}$ is a subset of the set $P$ of all sets of ordered pairs of real numbers, so $F$ is a set.

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