Subset of Naturals is Finite iff Bounded

Theorem
Let $X$ be a subset of the natural numbers $\N$.

Then $X$ is finite if and only if it is bounded.

Proof
Since the set of natural numbers is bounded below, $X$ is bounded iff $X$ is bounded above.

That is, iff there exists $p \in \N$ such that $x \le p$ for all $x \in X$.

Necessary Condition
Let $X$ be finite.

Then $X = \left\{{x_1, x_2, \ldots, x_n}\right\}$ for some $n \in \N$.

Let $p = x_1 + x_2 + \ldots + x_n$.

Then we see that $x \in X \implies x \le p$, hence $X$ is bounded.

Sufficient Condition
Now let $X \subseteq \N$ be bounded.

Then for some $p \in \N$, it is contained in the initial segment $\N_p = \left\{{0, 1, \ldots, p}\right\}$ of $\N$.

It follows from Subset of Finite Set is Finite that $X$ is finite.