infinite subset of an finite set?

Question

Is it possible to have a set of infinite cardinality as a subset of a set with a finite cardinality? It sounds counter-intuitive, but there are things in math that just are so. Can one definitely prove this using only basic axioms?
The main reason I asked this question is because the book Inverted World says there are infinite planetary bodies in a finite universe, and I wondered if this could be done with sets.

Answer 1

The proof is very intuitive (as you probably are feeling). But it can be written elaborately as follows, if you wish.

Your claim: For any finite set F, there exists an infinite subset I.

Try to prove: Let $F$ be a finite set defined as $F = \{f_1, f_2, \ldots , f_n\}$, where $n = 1, 2, \ldots$

Let $I$ be an infinite set defined as $I = \{i_1, i_2, \ldots, i_n, \ldots\}$, where n = 1, 2, ...

If I is a subset of F, then every element in I is also an element in F. If F contains finitely many elements, then only finitely many elements of I could belong to F.

However, I is infinite by definition, so clearly not all elements of I are contained in F.

Therefore, I is not a subset of F. This implies the claim is false.

Hence, for any finite set F, there does not exist an infinite subset I.

There is actually a proof you can probably find which does the same thing, just it takes a different angle: Prove that every subset of a finite set is finite. You can probably look this up somewhere!

I don't believe there are infinite planets in the universe. There are a large number, but it is not infinite. I don't believe anything in the universe is infinite, so there shouldn't be anything to reconcile here. Inverted World is sci-fi, so it's not even a theory. Just a nice tale!

Answer 2

By definition a set $B$ is a subset of $A$ iff every element of $B$ in $A$. So, the largest subset of a finite set $A$ has exactly as many elements as $A$, but no more.

Answer 3

"Length" is an inappropriate word here, partly because it's confusing and potentially ambiguous. One can say that the set of all numbers between $0$ and $1$ has finite "length", but it has infinitely many members. It is infinite in the sense usually used when talking about sets, i.e. it has infinitely many members. It is also bounded, both in the sense that it has upper and lower bounds ($1$ and $0$) and in the sense that there is an upper bound on the distances between its members (no two of them are more than a certain finite distance from each other. Among those members are $1/2,\ 1/3,\ 1/4,\ 1/5, \ldots$, and there are infinitely many of those within this set of finite "length".

Suppose we put the question this way: Can a finite set have an infinite subset? Or: Can a set with finitely many members have a subset with infinitely many members? Then the answer is clearly "no".

You should not say "infinite planets" if you mean "infinitely many planets". In standard usage in mathematics, "infinite planets" means "planets each one of which, by itself is infinite." If planet A is infinite and planet B is infinite, then those are infinite planets, but they are not infinitely many planets, since there are only two of them.