A variation of (0, 1) vs. R

The partial order of set cardinalities is defined thus: if there is an injection (1-1 function) from set $X$ to set $Y$ , then $|X| \leq |Y|$ .

The identity function serves as an injection from $\mathbb{R} \rightarrow \overline{\mathbb{R}}$ . Thus $|\mathbb{R}| \leq |\overline{\mathbb{R}}|$

For the opposition direction, define $f: \overline{\mathbb{R}} \rightarrow \mathbb{R}$ by

$f(-\infty) = -\frac{\pi}{2},$ $f(\infty) = \frac{\pi}{2}$ , and $f(x) = \tan^{-1}(x)$ for $x \in \mathbb{R}$ .

$f$ is a 1-1 function from $\overline{\mathbb{R}}$ to the interval $[-\frac{\pi}{2},\frac{\pi}{2}]$ . Thus

$|\overline{\mathbb{R}}| \leq |\mathbb{R}|$ , and since the inequality holds in both directions, $|\mathbb{R}| = |\overline{\mathbb{R}}|$

This problem can be solved by using the fact that the cardinality of the union of disjoint sets equals the sum of the sets' individual cardinalities. Therefore, the cardinality of the set of extended reals is $|\R\cup\{-\infty, +\infty\}|=|\R|+|\{-\infty, +\infty\}|$ . The cardinality of the real numbers can be denoted by $\mathfrak{c}$ , and the cardinality of the latter set is 2. By transfinite cardinal arithmetic, $\mathfrak{c}+2=\mathfrak{c}$ . But $\mathfrak{c}$ is defined as the cardinality of the real numbers! Therefore, $\fbox{they have the same cardinality}$ .