# Set with no elements!?

Consider a set defined on the universe of integers. It contains a large number of subsets. These subsets include the empty set and sets which contain multiple integer elements. However our set does not directly contain any integers.

Is it correct to say that our set does not contain any elements (despite the large number of subsets it contains)?

After all, these subsets are not elements but simply subsets. And their elements are not elements of our set, but merely elements of the subsets.

The set is not defined on the "universe of integers". Sets defined on the "universe of integers" have integers as elements. This is a set defined on the "universe of sets of integers", namely its elements are sets of integers. Yes, sets can be elements of other sets as well.

More correctly we should say that we consider a subset of the integers, or a subset of the power set of the set of integers.

As for elements, since if you only consider integers in your universe, a set of sets of integers is not a subset of your universal set, the question is essentially meaningless. From a broader mathematical standpoint, elementhood is relative to a model of set theory, and in models of set theory we usually consider many more objects than just integers, for example we consider sets of integers as well. Since you specifically state that the empty set is an element of your set, it is not without elements.

You're speaking of a set of sets. In this case, I believe that what you call subsets are in fact elements of the set of sets. That is, a set can be an element of a set of sets.

• Are they only elements of the set or are they both elements and subsets? – hb20007 Nov 5 '14 at 21:54
• I would say they are just elements. To be a subset of the set of sets, you'd have to have a set of those elements. Have a look at en.wikipedia.org/wiki/Family_of_sets – Shane Nov 5 '14 at 21:55
• Ah, because the set is actually defined on the universe of sets of integers, as Asaf pointed out. I get it now. Thanks, you two! – hb20007 Nov 5 '14 at 22:03
• Yes indeedy. Since Asaf's answer is more complete, you should probably accept his. – Shane Nov 5 '14 at 22:04
• I was going to pick yours because you answered first and also provided a link. But okay, so you say, I'll accept Assaf's! – hb20007 Nov 5 '14 at 22:06

No that is not right.

Think of your set as a power set of a set $A$. The elements of this powerset are subsets of $A$, but the elements of the powerset are not elements ansich of $A$.

So your set does in fact contain elements, however the elements are subsets of another set (which contains the elements, which make up the subsets)