Let $A$ be an infinite set.

Then, we can construct an injective function $f:\omega \rightarrow A$.

But how do i construct this via orginal statement of $AC_\omega$? (i.e. $\forall countable X, [\emptyset \notin X \Rightarrow \exists f:X\rightarrow \bigcup X \forall A\in X, f(A)\in A$)

So my question is:

(1) how do i prove that every infinite set has a countable subset?

(2) how do i write down a situation such as: "after choosing $x_1,...,x_k$, choosing $x_k$ satisfying a given condition. Continue this and form a sequence"


2 Answers 2


You need to find a countable family of non-empty sets that you can choose from.

For the first statement, the answer would require you to find a countable family of subsets which are non-empty. For example $A_n=\{A\subseteq X\mid A\text{ has exactly }n\text{ elements}\}$ is non-empty for $n\in\omega$, because $X$ is assumed to be infinite.

Next we use $\sf AC_\omega$ to claim that there exists $f$ whose domain is $\omega$ such that $f(n)\in A_n$ for every $n>0$. Now we want to claim that the union, $\bigcup\{f(n)\mid n\in\omega\}$ is countably infinite, but this requires us to step through another argument, why is the countable union of finite sets countable. I will let you figure this one out, but let me give you the general hint:

Suppose that $A_n$, for $n\in\omega$ is a family of countable sets, let $F_n$ be the functions of all injections from $A_n$ into $\omega$. Choose $f_n$ from $A_n$, and map the union $\bigcup A_n$ into $\omega\times\omega$, by mapping $A_n$ into $\{n\}\times\omega$, using $f_n$ (note that we didn't assume the $A_n$'s are disjoint, and we don't need to, but this requires another small argument).

The second question, requires a choice principle which is strictly stronger than $\sf AC_\omega$. Namely, the choice of $x_{n+1}$ depends on the choice of $\{x_0,\ldots,x_n\}$. This principle is called Dependent Choice, or $\sf DC$ (or $\sf DC_\omega$ sometimes), and it is formulated in many different ways. One of them is the following:

Suppose that $X$ is a non-empty set and $R$ is a binary relation on $X$ such that $\operatorname{dom}(R)=X$. Then there exists a function $f\colon\omega\to X$ such that $f(n)\mathrel{R}f(n+1)$ for all $n$.

The principle $\sf DC$ is well investigated, it is strictly stronger than $\sf AC_\omega$, and is implied by $\sf AC$ itself, of course. This is essentially a principle allowing us to extend inductive definitions to the existence of a sequence satisfying the conditions we want. So often in contexts where the full axiom of choice is present, because it is so easy to use, we use $\sf DC$ to prove results which only require $\sf AC_\omega$.

Some reading material on these choice principles and finite-ness:

Online discussions:

  1. Infinite Set is Disjoint Union of Two Infinite Sets
  2. Equivalent of the countable axiom of choice?
  3. Books on Axiom of Dependent Choices?
  4. Stronger than ZF, weaker than ZFC
  5. All naturals are T-finite, all finite sets are T-finite

Research papers:

You may want to check out the references in the above links, but additionally these may come in handy.

  1. Truss, John. "Classes of Dedekind finite cardinals." Fundamenta Mathematicae vol. 84.3 (1974): 187-208.
  2. Herrlich, Horst. "The finite and the infinite." Appl. Categ. Structures 19 (2011), no. 2, 455–468.
  • $\begingroup$ Could you tell me some useful links about $AC_\omega$ and $DC$? I really want to know 'boundary' of these two. For example, dedekind finite and taraki finite are the same if i assume $AC_\omega$? $\endgroup$
    – Jj-
    Sep 18, 2013 at 14:51
  • $\begingroup$ Well, there are a lot of fun questions on this site about these principles, and several papers. I can't right now, but in two or three hours I will have better access to everything and I'll add references to my answer. $\endgroup$
    – Asaf Karagila
    Sep 18, 2013 at 14:56
  • $\begingroup$ To your example, by the way, one can show that Tarski finite is always Dedekind finite. The other implication need not be true without some choice. However with countable choice every Dedekind finite is finite, so all the classes of finite sets are equal. $\endgroup$
    – Asaf Karagila
    Sep 18, 2013 at 14:58
  • $\begingroup$ Thank you! I also want to know whether $AC_\omega$ proves there does not exists a set $A$ such that for all finite ordinal $n\preceq A$ and $A\preceq \aleph_0$ but $A$ is not countable and etc :) $\endgroup$
    – Jj-
    Sep 18, 2013 at 15:16
  • $\begingroup$ If $A\preceq\aleph_0$ then either $A$ is finite or $A\sim\omega$. You can prove that without using the axiom of choice at all. Simply from the assumption $A\preceq\aleph_0$ which essentially means that there is an injection from $A$ into $\omega$, and therefore there is an injection from $A$ into an initial segment of $\omega$. If it's a proper initial segment then $A$ is finite. If not then $A$ is countably infinite. $\endgroup$
    – Asaf Karagila
    Sep 18, 2013 at 16:52

Here's an alternative proof that countable choice implies that every infinite set has a countably infinite subset. Assume countable choice, and let $A$ be an infinite set. For each natural number $n$ (identified, as usual, with the set $\{0,1,\dots,n-1\}$), let $S_n$ be the set of all one-to-one maps from $n$ into $A$. The assumption that $A$ is infinite implies (by induction on $n$) that $S_n\neq\varnothing$ for all $n$. So by countable choice, there is a sequence $(s_n)_{n\in\omega}$ such that $s_n\in S_n$ for al $n$. Now build an infinite sequence of distinct elements of $A$ as the concatenation of sequences $t_n$, where $t_n$ consists of the elements in the range of $s_n$ that are not in the range of any $s_k$ with $k<n$; these elements are to be listed in $t_n$ in the same order as in $s_n$. Some of the $t_n$ might be empty, but there are infinitely many $n$ for which $t_n$ is nonempty. (Proof: Otherwise, there would be an $m$ such that all $t_n$ for $n>m$ are empty, which means the ranges of all $s_n$'s are included in the union of the ranges of the $s_n$'s for $n\leq m$. But that union is finite and once $n$ is larger than the cardinality of that union, the range of $s_n$ has too many elements to be included in that union.) So the concatenation of all the $t_n$'s is an infinite sequence of distinct elements of $A$.

  • $\begingroup$ (+1) This the advantage of not using "countable union of countable sets is countable", as in the other answer. $\endgroup$ Feb 16, 2022 at 12:52

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