some elementary questions about cardinality

Question

I know little about set theory and while reading some Algebra proof I had difficulty on some details. So my questions are :

1. If $X$ is an infinite set and $Y$ is the set of all finite subsets of $X$, they have the same cardinality. How can I prove it ?
2. If $X$ is infinite, then it has the same cardinality of the product $X \times \mathbb{N}$, where $\mathbb{N}$ is the set of natural numbers. This seems pretty obvious, specially if you think separately when $X$ is countable and when it's not, but I cant prove this in a rigorous way.
3. Consider only infinite sets now. I know that if $X$ is countable, then it has the same cardinality as $X \times X$. Is this true in general (I mean, I know there is bijection between $\Bbb R$ and $\Bbb R^2$, but probably there is a counterexample that proves this is wrong for an arbitrary infinite set, or if not, I cannot prove it either) ?

Thank you so much for the help !

Answer 1

All these, and more, depend on an axiom called The Axiom of Choice. This axiom enforces infinite set to be very well-behaved, and the answer on all your questions become "yes".

If we do not assume the axiom of choice then the answers change, namely some of them become "sometimes" and others "no".

None of these propositions are trivial, especially without some background in set theory. It is even less trivial if you are not used to proving things from axiomatic systems like ZFC.

Let me answer your questions backwards.

• The axiom of choice is equivalent to the assertion "For every infinite $X$, $X\times X$ has the same cardinality as $X$". In particular this means that if we assume the axiom of choice then this is true for every infinite set, but if we assume that this axiom does not hold, then we are guaranteed a counterexample to exist.

• The second question regarding $X\times\mathbb N$, again if we assume the axiom of choice the answer is "yes". One of the consequences of this axiom is that if $X$ is infinite then $X$ has a countably infinite subset. Combine with the above we have that $$|X|\leq|X\times\mathbb N|\leq|X\times X|=|X|$$ Therefore equality ensues.

• This again amounts to the axiom of choice and the first answer. Assuming that $X\times X$ has the same cardinality as $X$, we can prove by induction that $|X^n|=|X|$ , and by the second answer we have that $|\bigcup X^n|=|X\times\mathbb N|=|X|$ again.

  This is not exactly what we are looking for because this is the set of sequences rather than subsets, but now we can simply send every sequence $\langle a_i\mid i=0,\ldots,k\rangle$ to the set $\{a_i\mid i=0,\ldots,k\}$. Observe that in the sequences there might be repetitions which are removed when we take the set, but we still can get every finite subset.

  Well, this is a surjection, so it does not quite yet prove that the set of all finite subsets of $X$ has the same cardinality as $X$, but another wonderful property following from the axiom of choice is that we can inverse surjective functions, namely there is an injection from the set of finite subsets into the set of finite sequences, and this is what we wanted.

One last remark is that the counterexamples without the axiom of choice are not very natural, at least for a modern person. However they exist and have their place in mathematics.

Answer 2

The results described below freely use the Axiom of Choice.

Theorem: If $X$ and $Y$ are infinite sets, then the cardinality of $X \times Y$ is the maximum of the cardinality of $X$ and the cardinality of $Y$. For a proof see e.g. Theorem 8 and the following discussion in these notes.

This result answers your third question and also your second question, because every infinite set contains a countably infinite subset.

As for your first question: one way to do it is to write the set $Y$ of all finite subsets of your infinite set $X$ as a countable union of sets $Y_n$ for $n \in \mathbb{N}$, where each $Y_n$ is the set of subsets of $X$ having at most $n$ elements. There is a natural surjection $X^n \rightarrow Y_n$ which shows that $\# Y_n \leq \# X^n = \# X$. Thus $Y$ is a countable union of sets each having cardinality at most $X$, so $Y$ has cardinality at most $X$ (you can think of this in terms of $\# X \times \# \mathbb{N} = \# X$, for instance). On the other hand, the subset $Y_1$ of one element subsets of $X$ is naturally in bijection with $X$ itself, so also $\# X \leq \# Y$. It follows from Cantor-Bernstein that $\# X = \# Y$.