Every infinite subset of a countable set is countable.

Question

Here is the proof I tried to weave while trying to prove this theorem:

Theorem. Every infinite subset of a countable set is countable.

Proof. Let $A$ be a countable set and $E\subset A$ be infinite. Then $A\thicksim\mathbb{N}$. This implies that there is a sequence $\left\{x_{n}\right\}_{n\in\mathbb{N}}$ where $x_{n}\in A$. Construct a sequence $\left\{n_{k}\right\}_{k\in\mathbb{N}}$ where $x_{n_{k}}\in A$, $n_{1}$ is the smallest positive integer such that $x_{n_{1}}\in A$, and $n_{k+1}>n_{k}$. Then

$$ E=\bigcup_{k=1}^{\infty}x_{n_{k}}, $$

and $E$ is countable, because $E\thicksim\mathbb{N}$. $\blacksquare$

Is it convincing?

Answer 1

I think some things can be written in a clearer manner. First, I would change

This implies that there is a sequence $\left\{x_{n}\right\}_{n\in\mathbb{N}}$ where $x_{n}\in A$.

For

This implies we can write $A=\{x_n:n\geqslant 1\}$

or

Let $\{x_n:n\geqslant 1\}$ be an enumeration of $A$.

Then, I would say

If $S$ is a subset of the natural numbers, let $\min S$ denote least element of $S$. Define $S_1=\{k:x_k\in E\}$. $S_2=S_1\setminus \{\min S_1\}$ and in general $$S_{n+1}=S_{n}\setminus\{\min S_1,\ldots,\min S_{n}\}$$ Then define $n_k=\mathscr \min S_k$

I guess the idea is clear: consider the set of subscripts such that $x_k\in E$. By the well ordering of the natural numbers, we can extract a sequence $n_k$ such that $$n_1<n_2<\cdots\\E=\{x_{n_k}:k\geqslant 1\}$$

by considering the first subscript with $x_k\in E$ removing this one from the list and looking at the new first subscript (our second in the list) and so on. Some details should be addressed

$(1)$ The set $S_{n}$ is never empty. Reason: Since $E\subseteq A$; $S_1$ is not empty. Moreover, $E$ is by assumption infinite, thus removing one element every time cannot exhaust it.

$(2)$ The construction exhausts the elements of $E$ -- that is, it is a surjection. Reason: Pick $m$ such that $x_m\in E$. We need to find $k$ such that $n_k=m$. Consider the finite set $\{x_1,\dots,x_m\}$. Keeping the order, remove all elements such that $x_i\notin E$. We're left with a finite set, and it must be the case $\{x_{n_1},\dots,x_{n_k}\}$ for some $k$, and $n_k=m$, by definition of the $n_k$.

$(3)$ The construction is an injection. Reason: By construction, $n_k\neq n_j$ if $j\neq k$ for if $j>k$ then $n_k\notin S_{j}$.

Conclusion We obtain an bijection of $E$ with an infinite subset $F$ of $\Bbb N$. Thus $E\simeq F\simeq \Bbb N$, that is $E\simeq \Bbb N$.

Answer 2

It seems correct, up to a typo between $A$ and $E$, and braces missing in the union at the end (in $\bigcup\{x_{n_k}\}$ Also, you might want to define $n_{k+1}$ as $$n_{k+1} = \inf\{ i > n_k \mid x_i \in E \}$$

Using the (equivalent, and pretty much identical up to notations sequence/function) definition of a countable set:

A set $A$ is called countable if there exists an injective function $f$ from $A$ to $\mathbb{N}$

you can also say that if $f\colon A \to \mathbb{N}$ is such an injective function, then in particular any restriction $f_{|_E}$ (for $E\subseteq A$) is also an injection, and thus $E$ is countable.

Answer 3

Let $A$ be a countably infinite set and $E\subset A$ be an infinite subset. Then $E$ is countably infinite.

The OP is using the 'can be bijectively enumerated' definition for a countably infinite set. Their proof concludes by writing $E$ as a union of a family of singletons indexed by $\Bbb N$.

Writing out a precise/formal proof that builds the OP's subsequence can be broken into parts, each of which is easily digestible.

Let $(x_{n})_{\, n\in\mathbb{N}}$ be a bijective enumeration of $A$.

Lemma 1: We can associate to any finite subset $F$ of $E$ with $k$ elements another subset $\Gamma(F)$ of $E$ satisfying

$\tag 1 F \subset \Gamma(F)$ $\tag 2 |\Gamma(F)| = k + 1$

Proof
Since $F$ is finite and $E$ is infinite, the set $S = \{m \in \mathbb N \, | \, x_m \notin F\}$ is nonempty. Let $\gamma(F)$ be the least integer in $S$ and define

$$ \Gamma(F) = F \cup \{ x_{ \gamma(F) } \}$$ $\blacksquare$

Using recursion and lemma 1 we can prove the following:

Proposition 2: There exists a family of sets $(F_n)_{\, n \ge 1}$ satisfying for every integer $n \ge 1$,

$\tag 3 F_n \subset E$ $\tag 4 F_n \subset F_{n+1}$ $\tag 5 |F_n| = n$ $\tag 6 E = \bigcup_{n \ge 1} F_n$

The OP can now 'take those singleton accretions' to define the bijective enumeration $(y_{n})_{\, n \ge 1}$ of $E$.