Why $\aleph_0$ and $c$ are unequal transfinite cardinals?

Question

I recently started to learn about set theory and cardinals, and in the book that I'm using says that since $\mathbb{Q}$ is countable and $\mathbb{R}$ is uncountable, we have $\aleph_0<c$, where $\aleph_0$ is the cardinality of $\mathbb{N}$ (and also $\mathbb{Q}$) and $c$ is the cardinal number of $\mathbb{R}$.

Now, my problem is with the next sentence that follows from the theorem:

Theorem: Let $S$ be a set. If $S$ is finite, then $|S|<\aleph_0$.

The book says:

Thus Theorem implies that $\aleph_0$ and $c$ are unequal transfinite cardinals.

and I don't understand why. The theorem is about finite sets and $\mathbb{Q}$ and $\mathbb{R}$ are infinite, how are the two statements related?

Answer 1

Here is what the book says:

It is customary to denote the cardinal number of $\Bbb R$ by $\mathcal{c}$, for continuum. Since $\Bbb Q\subseteq \Bbb R$, we have $\aleph_0 \le \mathcal{c}$. In fact, since $\Bbb Q$ is countable and $\Bbb R$ is uncountable, we have $\aleph_0<\mathcal{c}$. Thus Theorem 8.15(e) implies that $\aleph_0$ and $\mathcal{c}$ are unequal transfinite cardinals.

So the fact that $\aleph_0$ and $\mathcal{c}$ are unequal is already implied by $\aleph_0<\mathcal{c}$. What Theorem 8.15(e) tells us is that $\aleph_0$ and $\mathcal{c}$ are transfinite.