Countable Set equals Range of Sequence

== Theorem ==
Let $S$ be a set.
Then $S$ is countable there exists a sequence $\sequence {s_i}_{i \mathop \in N}$ where $N$ is a subset of $\N$ such that $S$ equals the range of $\sequence {s_i}_{i \mathop \in N}$.
== Proof ==
=== Necessary Condition ===
Assume that $S$ is countable.
We need to prove that there exists a sequence $\sequence {s_i}_{i \mathop \in N}$, $N \subseteq \N$, such that $S$ equals the range of $\sequence {s_i}_{i \mathop \in N}$.
The range of $\sequence {s_i}_{i \mathop \in N}$ is defined as $\set {s_i: i \in N}$.
;Case 1. $S$ is empty.
Empty Set is Countable confirms that $S$ is countable.
Define the empty sequence as $\sequence {s_i}_{i \mathop \in \O}$.
The range of the empty sequence is $\set {s_i: i \in \O} = \O$.
$S$ equals the range of the empty sequence as $S$ equals $\O$.
This finishes the argument for this case.
;Case $2$. $S$ is finite and not empty.
Let $n \in \N_{>0}$ be the number of elements of $S$.
Define a sequence $\sequence {s_i}_{1 \mathop \le \mathop i \mathop \le \mathop n}$ by going through every element of S:
:Let $s_1$ be any element of $S$.
:Let $s_2$ be any element of $S \setminus \set {s_1}$.
:...
:Let $s_n$ be the single element of $S \setminus \set {s_i: 1 \le i \le n - 1}$.
From this definition follows that every element of $S$ equals some $s_i$, $1 \le i \le n$.
So, $S \subseteq \set {s_i: 1 \le i \le n}$.
Every number $s_i$, $1 \le i \le n$, equals some element of $S$.
So, $\set {s_i: 1 \le i \le n} \subseteq S$.
Accordingly, $S = \set {s_i: 1 \le i \le n}$.
By the definition of range, $\set {s_i: 1 \le i \le n}$ equals the range of the sequence $\sequence {s_i}_{1 \mathop \le \mathop i \mathop \le \mathop n}$.
Therefore, $S$ equals the range of $\sequence {s_i}_{1 \mathop \le \mathop i \mathop \le \mathop n}$.
This finishes the proof for case 2.
;Case $3$. $S$ is infinite.
Since $S$ is countable and infinite, $S$ is countably infinite.
By the definition of countably infinite, $S$ can be written:
:$\set {t_0, t_1, \ldots, t_n, \ldots}$
where $n$ runs over all the natural numbers, $\N$.
In other words, we have:
:$S = \set {t_i: i \in \N}$
We intend to produce a sequence from $S$.
We start with the empty sequence.
Using $\N$, the set to which $S$ is equivalent, we put all the elements of $S$ one by one into the sequence.
The result is $\sequence {t_i}_{i \mathop \in \N}$.
The range of $\sequence {t_i}_{i \mathop \in \N}$ is $\set {t_i: i \in \N}$, which equals $S$.
In other words, $S$ is the range of the sequence $\sequence {t_i}_{i \mathop \in \N}$.
This finishes the necessary part of the proof.
=== Sufficient Condition ===
Assume that there exists a sequence $\sequence {s_i}_{i \mathop \in N}$ where $N$ is a subset of $\N$ such that $S$ equals the range of that sequence.
We need to prove that $S$ is countable.
;Case 1. $S$ is finite.
By the definition of countable $S$ is countable.
;Case 2. $S$ is infinite.
We know that $S$ equals the range of the sequence $\sequence {s_i}_{i \mathop \in N}$ where $N$ is an infinite subset of $\N$.
The range of $\sequence {s_i}_{i \mathop \in N}$ is defined as $\set {s_i: i \in N}$, so we have
:$S = \set {s_i: i \in N}$
We need to prove that $S$ is countably infinite.
That a set is countably infinite, means that it is of the form:
:$\set {t_0, t_1, \ldots, t_n, \ldots}$
where $n$ runs over all the natural numbers, $\N$.
Accordingly, we need to show that a set $\set {t_i: i \in \N}$ exists such that:
:$S = \set {t_i: i \in \N}$
Assume that $N = \N$.
Then $S$ is countably infinite by definition.
The only other alternative is $N \subset \N$.
Define:
:$t_i = s_i$ for every $i$ in $N$
:$t_i = s_1$ for every $i$ in $\N \setminus N$
Thus, $t_i$ is defined for every $i$ in $\N$.
We have:
All in all, $S = \set {t_i: i \in \N}$.
Therefore, $S$ is countably infinite and thus countable.
This finishes the last part of the proof.
 
== Sources ==
* : Ch. $1$: Section $6$, Definition
Category:Countable Sets