# Cardinality of the real numbers [duplicate]

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Why is cardinality the of real numbers $2^{\aleph_0}$? While I know that real numbers can be constructed using sets of natural numbers, that solely does not mean that the real numbers has cardinality $2^{\aleph_0}$. So, what makes the real numbers cardinality $2^{\aleph_0}$?

I can't find why it is like this in my textbook...

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## marked as duplicate by William, Cookie, Normal Human, Jack D'Aurizio, Noah SnyderAug 4 '14 at 6:02

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Cantor set.${}{}{}$ – Asaf Karagila Jun 2 '12 at 8:32
Binary expressions – Henry Jun 2 '12 at 8:56

## 1 Answer

Use the Cantor–Bernstein–Schroeder theorem:

Choose $x\in \mathbb{R}$, and let $\{b_i\}_{i=1}^{\infty}$ be the (shortest) binary expansion of $\frac{1}{2}+\frac{\arctan x}{\pi}$ (which is in $(0,1)$). Then define $\phi(x) = \{ i | b_i = 1 \}$. This establishes an injective map $\phi:\mathbb{R}\to 2^{\mathbb{N}}$.

To go the other way, suppose $A \subset \mathbb{N}$. Then let $t_i = 1_A(i)$ (indicator function), and define $\eta(A) = \sum_{i=1}^{\infty}\frac{t_i}{3^i}$. This establishes an injective map $\eta : 2^{\mathbb{N}} \to \mathbb{R}$.

The desired result follows from the Cantor–Bernstein–Schroeder theorem.

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