=== 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. {{qed}} === 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: {{begin-eqn}} {{eqn | l = \set {t_i: i \in \N}       | r = \set {t_i: i \in N} \cup \set {t_i: i \in \N \setminus N}       | c = as $\N = N \cup \paren {\N \setminus N}$ }} {{eqn | r = \set {s_i: i \in N} \cup \set {t_i: i \in \N \setminus N}       | c = as $s_i = t_i$ for every $i$ in $N$ }} {{eqn | r = S \cup \set {t_i: i \in \N \setminus N}       | c = by the definition of $S$ }} {{eqn | r = S \cup \set {s_1}       | c = as $t_i = s_1$ for every $i$ in $\N \setminus N$ }} {{eqn | r = S       | c = as $\set {s_1} \subset S$ as $s_1 \in S$ }} {{end-eqn}} 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. {{qed}}