Under this set of definitions, prove that any subset of a countable set is countable

Question

Given that:

Two sets A and B are equivalent if there exists a bijection from A to B.

A set is finite if it is empty or is equivalent to the set {1,2,...,n} for some natural number n. Otherwise it is infinite.

A set is denumerable if it is equivalent to the set of natural numbers N.

A set is countable if it is finite or denumerable.

I understand that there are some questions in math stackexchange about proving any subset of a countable set is countable, which has already been answered. But I am just having troubles with proving this statement using the above definitions, which are sort of different from the common definitions.

Could someone give me a hand, please? Thanks so much.

Answer 1

Let $X$ be a denumerable set and $Y$ a subset of $X$, there exists an injection $f:X\rightarrow N$. $f(Y)$ has a smallest element $f(x_0)$, define $g(x_0)=0$, suppose defined $g(x_0),...,g(x_n)$, $f(Y-\{x_0,...,x_n\})$ has a smallest element, $f(x_{n+1})$, defined $g(x_{n+1})=n+1$.

1. There exists $n$ such that $Y-\{x_0,...,x_n\}$ is empty we deduce $Y=\{x_0,...,x_n\}$.

2. Suppose that $Y$ is infinite and $Y$ distinct of $\bigcup_n\{x_0,...,x_n\}$. Let $y\in Y-\bigcup_n\{x_0,...,x_n\}$. $f(y)=f(x_{f(y)})$ since $f$ is injective, $x_{f(y)}=y$ contradiction. Thus $Y=\{x_0,...,x_n,...\}$ and $g$ is a bijection.