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This is an exercise in Naive Set Theory by P. R. Halmos.

If $\text{card }A=a$, what is the cardinal number of the set of all one-to-one mappings of $A$ onto itself? What is the cardinal number of the set of all countably infinite subsets of $A$?

It is easy to see that, for the first question, the cardinal number is less than or equal to $a^a$, and for any finite set the cardinal number is $a!$; for the second one, the cardinal number is less than or equal to $2^a$, and for any finite set the cardinal number is $0$.
I guess for infinite sets the equality holds in both questions. Is that correct? Can anyone give some suggestions on what to do next?
Edit: As is suggested by Arthur, the second cardinal number is at most $a^\omega$.

Eclipse Sun
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    There are at most as many subsets that there are functions $\omega\to A$, so the cardinality is at most $a^\omega$. – Arthur Nov 21 '14 at 15:28
  • @DietrichBurde When I say two cardinal numbers are equal, I mean that the correspondent ordinal numbers are equivalent. – Eclipse Sun Nov 21 '14 at 15:38
  • Are you assuming the axiom of choice? In either case both these questions has been answered several times before. This should be closed as a duplicate (of at least one), but with the bounty this is impossible. – Asaf Karagila Nov 24 '14 at 09:24
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    This question shows there is a maximal number of 1-1 functions since those which are bijections already have that cardinality; and this question shows that indeed $a^\omega$ is the right cardinal for an infinite set, and it is $0$ for a finite set, as you observed. Of course, both assuming choice, otherwise all hell breaks loose. – Asaf Karagila Nov 24 '14 at 09:32
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    This question appears to be off-topic because it contains two largely unrelated questions which should be asked separately. – user642796 Dec 02 '14 at 17:40
  • @ArthurF: Yes; but also, they shouldn't be, because they are both duplicates! :-) – Asaf Karagila Dec 02 '14 at 18:36
  • @ArthurFischer Actually, this is one exercise but not two. I thought they were interrelated, but now it appears to be not. – Eclipse Sun Dec 03 '14 at 09:53

2 Answers2

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If $|A|=a$ is infinite then the cardinal number of the set of all countably infinite subsets of $A$ is $a^\omega$.

The upper case is trivial because every subset can be represented by an ordered sequence and there are $|A|^\omega$ sequences.

Due to $a^2=a$, the set $\mathbb{Z}\times A$ is equivalent to $A$, so there is a bijection $\varphi:A\to (\mathbb{Z}\times A)$. So the set $A$ can be partitioned into infinitely many disjoint subsets, all of cardinality $a$: $$ A=\bigcup_{n=1}^\infty A_n; \quad |A_1|=|A_2|=\ldots=|A|=a. $$ Now there are $a^\omega$ distinct subsets that contain exactly one element of every $A_n$.

Greg Martin
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G.Kós
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For the first question: Suppose that $A$ is infinite, since every function $f:A\rightarrow A$ is given by a subset of $A\times A$, and $a^2=a$, the cardinality of the set of all functions from $A$ to $A$ is at most $2^a$. In fact $a^a=2^a$ since $2\leq a$ implies $2^a\leq a^a\leq 2^a$.
Now we prove that the cardinality of all the bijective functions is $2^a$ and then we are done with the first question. Given any subset $S\subset A$ we choose any bijective function $f_S:A\setminus S\rightarrow A\setminus S$ without a fixed point (prove that there is at least one, by Zorn) and we extend this function to a bijective function from $A$ to $A$ as the identity in $S$. This proves that there is at least $2^a$ bijective functions from $A$ to $A$, so the equality holds.

Eclipse Sun
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matiasdata
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