# On Cardinalities of Finite Sets

In mathematics often when something "doesn't make sense", it turns out that it exists anyway.

For example we now know there are numbers whose squares are negative. There are geometric sets with fractional dimension, etc.

So if we have a finite set, "common sense" dictates that it's cardinality is some non-negative integer.

I'm wondering if mathematicians work with sets with negative cardinalities or non-integral rational number cardinalities (how about irrational, complex?) etc.

Or if to our knowledge these concepts do not make sense.

• I can't think of a meaningful definition of negative cardinality. But there is something to be said about 'negative order types'. For finite sets the negative and positive order types agree (they are isomorphic) and are equal to the cardinality. For infinite sets all three notions become distinct concepts. If you think about order types as structural cardinalities, this gives a positive answer to your question. – Stefan Mesken Mar 4 '18 at 21:47
• There's an obstacle that there's no way to subtract one set from another in a way that respects cardinality: for example, $\aleph_0 - \aleph_0$ could be seen as $\mathbb{N} \setminus \mathbb{N}$ or as $\mathbb{N} \setminus 2 \mathbb{N}$, giving two very different cardinalities. – Patrick Stevens Mar 4 '18 at 21:47
• In my reading, what you're after is a generalization of natural numbers, which arose first as cardinalities of sets. Well, negative, rational, etc. numbers themselves are just that in a way. If you insist to find 'set-like' objects with negative/real cardinalities, consider e.g. money or sets of apples that you can cut in any proportion.. By the way, one can work with weighted sets, which are just functions $X\to\Bbb R$ in a similar manner as multisets work (which may contain an element many times). – Berci Mar 4 '18 at 22:00
• mathoverflow.net/questions/136366/… – Asaf Karagila Mar 4 '18 at 22:22

What would be the disjoint union of a set of cardinality $-1$ and a set of cardinality $1$? Is it going to be empty? If not, then cardinality no longer obey some of the basic laws we expect it to obey.