Aleph Zero is less than Cardinality of Continuum

Theorem
$\aleph_0 < \mathfrak c$

where
 * $\aleph$ denotes the aleph mapping,
 * $\mathfrak c$ denotes continuum, the cardinality of real numbers.

Proof
By Continuum equals Cardinality of Power Set of Naturals:
 * $\mathfrak c = \left\vert{\mathcal P \left({\N}\right)}\right\vert$

where
 * $\mathcal P \left({\N}\right)$ denotes the power set of $\N$
 * $\left\vert{\mathcal P \left({\N}\right)}\right\vert$ denotes the cardinality of $\mathcal P \left({\N}\right)$.

By Cardinality of Set less than Cardinality of Power Set:
 * $\left\vert{\N}\right\vert < \left\vert{\mathcal P \left({\N}\right)}\right\vert$

Thus by Aleph Zero equals Cardinality of Naturals
 * $\aleph_0 < \mathfrak c$