The cardinality of ${\mathbb{R}}$

We already gave a name to $\char93 {\mathbb{R}}$, namely ${\mathfrak{c}}$, but in fact, this was unnecessary, since we'll soon see that $\char93 {\mathbb{R}}=2^{\aleph_0}$.

Theorem 7 (Typewriter Principle II)   Let $S$ be a set. If there is a way to label each element of $S$ with an infinite sequence of typewriter symbols (like a!b!c#d&$\cdots$) so that no two elements of $S$ are given the same label, then $\char93 S \le 2^{\aleph_0}$.

Proof. Assign each typewriter symbol a code of exactly 10 binary digits. Concatenating the codes in an infinite sequence of typewriter symbols yields an infinite sequence of binary digits. If we map each element of $S$ to the corresponding binary digit sequence, we get an injection $S \rightarrow \{0,1\}^{\mathbb{N}}$. Hence $\char93 S \le \char93 \{0,1\}^{\mathbb{N}}= 2^{\aleph_0}$. $\qedsymbol$

Theorem 8   We have $\char93 {\mathbb{R}}=2^{\aleph_0}$; in other words, ${\mathfrak{c}}=2^{\aleph_0}$.

Proof. Any real number can be labelled by its decimal expansion, which is an infinite sequence of typewriter symbols like

$$-386.589734957938798379579057\dots$$

so Typewriter Principle II implies that $\char93 {\mathbb{R}}\le 2^{\aleph_0}$. On the other hand there is an injection $\{0,1\}^{\mathbb{N}}\rightarrow {\mathbb{R}}$ sending each infinite sequence of 0's and $1$'s to the corresponding real number having those as the digits past the decimal point; for instance

$$1011101110\dots \quad \longrightarrow \quad .1011101110\dots.$$

This injection shows that $\char93 \{0,1\}^{\mathbb{N}}\le \char93 {\mathbb{R}}$; i.e., $2^{\aleph_0} \le \char93 {\mathbb{R}}$. Combining these shows that $\char93 {\mathbb{R}}=2^{\aleph_0}$. $\qedsymbol$

We showed earlier that the set $\overline{{\mathbb{Q}}}$ of algebraic numbers had size only $\aleph_0$, so the same is true for the real algebraic numbers. But $\aleph_0 < 2^{\aleph_0} =\char93 {\mathbb{R}}$, so this shows that at least some real numbers are transcendental!