Subset of Natural Numbers is Cofinal iff Infinite

Theorem
Consider the ordered set $\left({\N, \le}\right)$, where $\le$ is the usual ordering on the natural numbers.

Let $\Sigma \subseteq \N$.

Then $\Sigma$ is cofinal iff it is infinite.

Proof
From Rule of Transposition, we may replace the only if statement by its contrapositive.

Therefore, the following suffices:

Implication
Suppose $\Sigma$ is infinite.

Let $n \in \N$ be an arbitrary natural number.

Then as $\N_n = \{m \in \N: m \le n\}$ is a finite set, it can't be that $\Sigma \subseteq \N_n$ by Subset of Finite Set is Finite.

Hence, as $\le$ is a total ordering, there exists a $\sigma \in \Sigma$ such that $n \le \sigma$.

Hence, as $n$ was arbitrary, $\Sigma$ is cofinal.

Contrapositive Implication
Suppose now that $\Sigma$ is finite.

As $\le$ is a total ordering, $\Sigma$ has a maximal element; call it $N$.

Then as $N + 1 \not\le N$, transitivity of $\le$ and maximality of $N$ assure us that we have:


 * $\forall \sigma \in \Sigma: N + 1 \not\le \sigma$

Hence $\Sigma$ is not cofinal.