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Difference between revisions of "Uncountable"
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We give an indirect proof here. This is one of the most famous indirect proofs and was given by George Cantor.
 
We give an indirect proof here. This is one of the most famous indirect proofs and was given by George Cantor.
  
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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 binary expansion of each <math>\omega_i</math>. For diedic rationals, take only the expansion terminating in an infinite chain of zeros, so that we have a unique binary expansion for each element of <math>A</math>. Say <math>\omega_i=(.b_{i1}b_{i2}b_{i3}...)_2</math> for all <math>i</math>. Now construct <math>\omega=(.b_1b_2b_3...)_2</math> such that <math>b_i=1-b_{ii}</math> for all <math>i</math>. So <math>\omega</math> is different from <math>\omega_i</math> for all <math>i</math>. So <math>\omega\not\in A</math>. But <math>\omega\in\mathbb{R},0<\omega<1</math>, a'' contradiction''. So <math>\mathbb{R}</math> is uncountable.
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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 binary expansion of each <math>\omega_i</math>. For dyadic rationals, take only the expansion terminating in an infinite chain of zeros, so that we have a unique binary expansion for each element of <math>A</math>. Say <math>\omega_i=(.b_{i1}b_{i2}b_{i3}...)_2</math> for all <math>i</math>. Now construct <math>\omega=(.b_1b_2b_3...)_2</math> such that <math>b_i=1-b_{ii}</math> for all <math>i</math>. So <math>\omega</math> is different from <math>\omega_i</math> for all <math>i</math>. So <math>\omega\not\in A</math>. But <math>\omega\in\mathbb{R},0<\omega<1</math>, a'' contradiction''. So <math>\mathbb{R}</math> is uncountable.
  
 
==See Also==
 
==See Also==

Revision as of 19:14, 5 December 2007

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

A set $S$ is said to be uncountable if there is no injection $f:S\to\mathbb{N}$. A well-known 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 given by George 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 $A$. Consider the binary expansion of each $\omega_i$. For dyadic rationals, take only the expansion terminating in an infinite chain of zeros, so that we have a unique binary expansion for each element of $A$. Say $\omega_i=(.b_{i1}b_{i2}b_{i3}...)_2$ for all $i$. Now construct $\omega=(.b_1b_2b_3...)_2$ such that $b_i=1-b_{ii}$ for all $i$. So $\omega$ is different from $\omega_i$ for all $i$. So $\omega\not\in A$. But $\omega\in\mathbb{R},0<\omega<1$, a contradiction. So $\mathbb{R}$ is uncountable.

See Also

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