Difference between revisions of "Uncountable"

(Proof that \mathbb{R} is uncountable)
Line 5: Line 5:
 
We give an [[indirect proof]] here. This is one of the most famous indirect proofs and was first given by [[Georg Cantor]].
 
We give an [[indirect proof]] here. This is one of the most famous indirect proofs and was first given by [[Georg Cantor]].
  
Suppose that the set <math>A=\{x\in\mathbb{R}:0<x< 1\}</math> is countable. Let <math>\{\omega_1, \omega_2, \omega_3, ...\}</math> be any enumeration of the elements of <math>A</math>. Consider the [[decimal expansion]] of each <math>\omega_i</math>, say <math>\omega_i=0.b_{i1}b_{i2}b_{i3} \ldots</math> for all <math>i</math>.  Now construct a real number <math>\omega= 0.b_1b_2b_3 \ldots</math> by choosing the digit <math>b_i</math> so that it differs from <math>b_{ii}</math> by at least 3 and so that not every <math>b_i</math> is equal to 9 or 0.  It follows that <math>\omega</math> differs from <math>\omega_i</math> by at least <math>\frac{2}{10^i}</math>, so <math>\omega \neq \omega_i</math> for every <math>i</math> and <math>\omega \not \in A</math>.  However, <math>\omega</math> is clearly a real number between 0 and 1, a [[contradiction]].  Thus our assumption that <math>A</math> is countable must be false, and since <math>\mathbb{R} \supset A</math> we have that <math>\mathbb{R}</math> is uncountable.
+
Suppose that the set <math>A=\{x\in\mathbb{R}:0<x< 1\}</math> is countable. Let <math>\{\omega_1, \omega_2, \omega_3, ...\}</math> be any enumeration of the elements of <math>A</math> (in other words, take an injection <math>f: A \to \mathbb{N}</math>, and denote <math>\omega_i = f(i)</math>).
 +
 
 +
Consider the [[decimal expansion]] of each <math>\omega_i</math>, say <math>\omega_i=0.b_{i1}b_{i2}b_{i3} \ldots</math> for all <math>i</math>.  Now construct a real number <math>\omega= 0.b_1b_2b_3 \ldots</math>, by choosing the digit <math>b_i</math> so that it differs from <math>b_{ii}</math> by at least 3 and so that <math>b_i</math> is never equal to 9 or 0.  It follows that <math>\omega</math> differs from <math>\omega_i</math> by at least <math>\frac{2}{10^i}</math>, so <math>\omega \neq \omega_i</math> for every <math>i</math>.  Thus, <math>\omega \not \in A</math>.  However, <math>\omega</math> is clearly a real number between 0 and 1, a [[contradiction]].  Thus our assumption that <math>A</math> is countable must be false, and since <math>\mathbb{R} \supset A</math> we have that <math>\mathbb{R}</math> is uncountable.
  
 
==See Also==
 
==See Also==

Revision as of 09:03, 13 August 2011

A set $S$ is said to be uncountable if there is no injection $f:S\to\mathbb{N}$. Every set that is not uncountable is either finite or countably infinite. The most common example of an uncountable set is the set of real numbers $\mathbb{R}$.

Proof that $\mathbb{R}$ is uncountable

We give an indirect proof here. This is one of the most famous indirect proofs and was first given by Georg Cantor.

Suppose that the set $A=\{x\in\mathbb{R}:0<x< 1\}$ is countable. Let $\{\omega_1, \omega_2, \omega_3, ...\}$ be any enumeration of the elements of $A$ (in other words, take an injection $f: A \to \mathbb{N}$, and denote $\omega_i = f(i)$).

Consider the decimal expansion of each $\omega_i$, say $\omega_i=0.b_{i1}b_{i2}b_{i3} \ldots$ for all $i$. Now construct a real number $\omega= 0.b_1b_2b_3 \ldots$, by choosing the digit $b_i$ so that it differs from $b_{ii}$ by at least 3 and so that $b_i$ is never equal to 9 or 0. It follows that $\omega$ differs from $\omega_i$ by at least $\frac{2}{10^i}$, so $\omega \neq \omega_i$ for every $i$. Thus, $\omega \not \in A$. However, $\omega$ is clearly a real number between 0 and 1, a contradiction. Thus our assumption that $A$ is countable must be false, and since $\mathbb{R} \supset A$ we have that $\mathbb{R}$ is uncountable.

See Also


This article is a stub. Help us out by expanding it.