# 8.9: Cardinality of the Real Numbers

• Bob Dumas and John E. McCarthy
• University of Washington and Washington University in St. Louis

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We finished Chapter 6 with the unproved claim that the real numbers are uncountable. Now that we have a formal definition of the real numbers, we are ready to complete our investigation of the cardinality of $$\mathbb{R}$$. By Theorem $$6.11$$ the set of infinite decimal sequences is uncountable, with cardinality $$2^{\aleph_{0}}$$. We went on to claim that this had consequences for the cardinality of $$\mathbb{R}$$. We consider the related question of the cardinality of the closed unit interval $$[0,1]$$.

Proposition 8.16. $$|[0,1]|=|\mathbb{R}|$$.

Proof. Define $$f:[0, \infty) \rightarrow(1 / 2,1]$$ by $f(x)=\frac{1}{x+2}+1 / 2 \text {. }$ Then $$f$$ is an injection. Let $$\mathbb{R}^{-}$$be the negative real numbers, and define $$g: \mathbb{R}^{-} \rightarrow[0,1 / 2)$$ by $g(x)=\frac{-1}{x-2} .$ Then $$g$$ is an injection. Let $$h: \mathbb{R} \rightarrow[0,1]$$ be the union of the functions $$f$$ and $$g$$. Then $$h$$ is clearly an injection. The identity function on $$[0,1]$$ is an injection into $$\mathbb{R}$$. By the Schröder-Bernstein Theorem, $|[0,1]|=|\mathbb{R}| \text {. }$ We investigate the relationship between infinite decimal expansions (which are related to infinite decimal sequences) and the real numbers. We restrict our attention to infinite decimal expansions of numbers in the unit interval $$[0,1]$$.

DEFINITION. Infinite decimal expansion For all $$n \in \mathbb{N}^{+}$$, let $$a_{n} \in$$ $$\ulcorner 10\urcorner$$. Then $a_{1} a_{2} \ldots a_{n} \ldots$ is an infinite decimal expansion. Let $$s$$ be an infinite decimal expansion $$a_{1} a_{2} \ldots$$. For $$n \in \mathbb{N}$$, let $s_{n}:=a_{1} \ldots a_{n}=\sum_{k=1}^{n} a_{k} 10^{-k} .$ We want to associate infinite decimal expansions with real numbers (understood as Dedekind cuts). We interpret infinite decimal expansions as Cauchy sequences of rational numbers.

Let $$D$$ be the set of infinite decimal expansions, and let $$f: D \rightarrow \mathbb{R}$$ be defined by $f\left(. a_{1} \ldots\right)=\lim _{n \rightarrow \infty} s_{n} .$ The sequence $$\left\langle s_{n}\right\rangle$$ is a Cauchy sequence so it converges to a real number. Let $L:=\left\{x \in \mathbb{Q} \mid(\exists n \in \mathbb{N}) x<s_{n}\right\} .$ The set $$L$$ is a Dedekind cut and $$f(s)=L$$. That is $\lim _{n \rightarrow \infty} s_{n}=L .$ $$L$$ is the least upper bound of the set $$\left\{s_{n} \mid n \in \mathbb{N}\right\}$$. We can associate with every infinite decimal expansion a real number in the unit interval, and can thereby define a function $$f: D \rightarrow[0,1]$$. Is $$f$$ a surjection? That is, can every real number in the unit interval be realized as an infinite decimal expansion? Let $$x \in[0,1]$$. We define an increasing sequence of rational numbers converging to $$x$$. For $$n \in N^{+}$$, let $$s_{n}$$ be the largest decimal expansion to $$n$$ decimal places that is no greater than $$x$$. If $$n<m$$, then $$s_{n}$$ is a truncation of $$s_{m}$$. Let $s=\lim _{n \rightarrow \infty} s_{n} .$ Then $$f(s)=x$$. Therefore $$f$$ is a surjection onto $$[0,1]$$.

It would be ideal if $$f$$ were an injection, for it would follow that Dedekind cuts are just the infinite decimal expansions. However this is not true. Suppose that $s=. a_{1} \ldots a_{n} a_{n+1} \ldots$ where $$a_{n} \neq 9$$ and for all $$k>n, a_{k}=9$$. If $t=. a_{1} \ldots a_{n-1}\left(a_{n}+1\right) 000 \ldots$ then $f(s)=f(t) .$ If neither $$s$$ nor $$t$$ are infinite decimal expansions that terminate in repeating 9’s, and $$s<t$$, then there is some $$n$$ such that $$s<t_{n}$$. So the rational number $$\left(s_{n}+t_{n}\right) / 2$$ is in the Dedekind cut $$f(t)$$ and not in $$f(s)$$, so $$f(s) \neq f(t)$$. Therefore we have proved the following theorem.

THEOREM 8.17. Let $$D_{0}$$ be the set of infinite decimal expansions for numbers in the unit interval. Let $$f: D_{0} \rightarrow[0,1]$$ be defined by $f\left(. a_{1} a_{2} \ldots\right)=\lim _{n \rightarrow \infty} . a_{1} \ldots a_{n}=\sum_{k=1}^{\infty} a_{k} 10^{-k} .$ Then $$f$$ is a surjection. Moreover, two distinct decimal expansions are identified by $$f$$ iff one of them is of the form $$a_{1} a_{2} \ldots a_{n} 9999 \ldots$$ with $$a_{n} \neq 9$$ and the other is . $$a_{1} a_{2} \ldots\left(a_{n}+1\right) 000 \ldots .$$

COROLLARY 8.18. $$|[0,1]|=2^{\aleph_{0}}$$.

PROOF. By Lemma 8.17, Proposition $$6.15$$ and Theorem 6.11, $|[0,1]| \leq\left|D_{0}\right|=\left|\ulcorner 10\urcorner^{\mathbb{N}}\right|=2^{\aleph_{0}} .$ Let $$g:\ulcorner 2\urcorner^{\mathbb{N}+}\rightarrow [0,1] D_{0}$$ be defined by $g\left(\left\langle a_{n}\right\rangle\right)=. a_{1} a_{2} \ldots$ and $$h: D_{0} \rightarrow[0,1]$$ be defined as in the argument for Theorem 8.17. Then $$h \circ g:\ulcorner 2\urcorner^{\mathbb{N}} \rightarrow[0,1]$$ is an injection, and so $2^{\aleph_{0}} \leq|[0,1]| .$ By the Schröder-Bernstein Theorem, $|[0,1]|=2^{\aleph_{0}} .$ COROLLARY 8.19. $$|\mathbb{R}|=2^{\aleph_{0}}$$.

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