Prove that every infinite set has a countable subset. I just need help trying to create a proof that shows that an infinite set has a countable subset. Is it as simple as taking arbitrary values of the finite set and listing them in their own subset?
 A: Definition: The statement that a set $S$ is infinite means that if $N$ is a natural number then $S$ contains $N$ distinct elements.
[Note: If an infinite set is defined in this way, then it automatically follows that an infinite set minus a finite set is infinite.]
Suppose $S$ is infinite.
Since $1$ is a natural number, $S$ contains an element $x_1$.
Since $2$ is a natural number, $S$ contains an element $x_2$ distinct from $x_1$.
So there is a two-element subset $U_2=\{x_1,x_2\}$ of distinct elements of $S$.
For each $N\in\mathbb{N}$, $S$ contains an element distinct from each element in $U_N=\{x_1,x_2,\cdots,x_n\}$, so
define $U_{N+1}=U_N\cup\{x_{N+1}\}$ where $x_{N+1}$ is an element of $S$ distinct from each element of $U_N$.
Let $$U=\bigcup_{N\in\mathbb{N}}U_N$$
Then $U$ is a countable subset of $S$.
ADDENDUM
There is an issue which I glossed over when making this argument.
For each $N$ we know that there is a subset $V$ of $S$ containing $N+1$ elements of $S$. So $V$ contains an element, call it $x_{N+1}$, which is distinct from each element of $U_N=\left\{x_n,x_2,\cdots,x_N\right\}$. Let $U_{N+1}=\left\{x_n,x_2,\cdots,x_N,x_{N+1}\right\}$.
A: Looking over the comments between John Wayland Bales and fleablood, I thought I would take a stab at this 'iconoclast approach', working in the realm of intuitive set theory.
Definition: A set is $S$ finite if there exist a natural number $n \gt 0$ such that every mapping $f: \{1, 2, \dots, n\} \to S$, the mapping $f$ is not an injection.
For a finite set $S$ we can consider the smallest $n_0$ where there are no injections. We then can say that $S$ has $n_0 - 1$ elements.
Proposition 1: If a finite set $S$ has $m$ elements then $S$ is equinumerous with $\{1,2,\dots,m\}$.
Proof
Let $g$ be an injection of $\{1,2,\dots,m\}$ into $S$. This mapping is necessarily a surjection. Suppose, to get a contradiction, that an element $s_0 \in S$ is not in the range of $g$. Then we can extend $g$ to another injective function by defining $g(m+1) = s_0$, but this is absurd. $\quad \blacksquare$
A set $S$ is infinite if it is not finite. This means that for every $n \gt 0$ there exist an injective mapping from $\{1,2,\dots,n\}$ into $S$.
Considering Dan Christensen's answer to the OP's question, we introduce the following axiom:
A set $S$ is infinite if and only there exist a function $f: S \to S$ that is an injection but is not a surjection.
A: Definition: A set is infinite, if it can't be mapped one-to-one with an n-element set for any natural number n.
Lemma (can be proven using the principle of induction): If a set is infinite, then it has an n-element subset for every natural number n.
Proof:


*

*Let $X$ be any infinite set.

*By axiom of separation on $\omega \times P(P(X))$, there exists a function $f$ from non-zero natural numbers, where $f(n)$ is the set of all n-element subsets of $X$. (By lemma, $f(n)$ is non-empty for every n.)

*By axiom of countable choice, there exists a sequence of sets $A_1, A_2, \dots$, such that $A_n$ is an n-element subset of $X$.

*Let's define the following sequence of sets: $B_0=A_1$, $B_1=A_2 \setminus A_1$, $B_2=A_4 \setminus A_2 \setminus A_1$, ...; $B_n$ is the set of all elements of $A_{2^n}$, which are not an element of any $A_{2^m}$ where $m<n$ (any of $A_1$, $A_2$, $A_4$, ..., $A_{2^{n-1}}$). Observe that all these sets are finite, disjoint, and non-empty (the set $B_n$ has excluded at most $2^{n}-1$ different elements from $A_{2^n}$).

*By axiom of countable choice, there exists a set containing a single element from each $B_n$. This set is a countably infinite subset of $X$. QED.

A: Let A be finite and non-empty then it is countable, by definition.
If A is infinite then pick $a_1 \in A$. Then $A \setminus \{a_1\}$ is infinite. Pick $a_2 \in A \setminus \{a_1\}$. Then $A \setminus \{ a_1,a_2\}$ is infinite. Then the for each positive integer i, choose $a_i \in A \setminus \{ a_1, \dots ,a_{i-1} \}$. Hence the set S, defined by $a_i \in S$ $\iff$ i is a positive integer, is a countably infinite set hence countable.
In summary,  you choose distinct points in the infinite set and form a set that consists of these points. Note that any set that consists of distinct points only is countable as you can index the terms by the positive integers and hence define a trivial bijection from the set to the positive integers.
