Difference between revisions of "Uncountable"
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| − | {{ | + | A [[set]] <math>S</math> is said to be '''uncountable''' if there is no [[injection]] <math>f:S\to\mathbb{N}</math>. Assuming the [[Axiom of choice]], 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 number]]s <math>\mathbb{R}</math>. |
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| + | == Proof that <math>\mathbb{R}</math> is uncountable == | ||
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| + | We give an [[indirect proof]] here. This is one of the most famous indirect proofs and was first given by [[Georg 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> (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. | |
| − | ( | + | An alternative proof uses [[Cantor's Theorem]], which says that for all sets <math>S</math>, we have <math>|S|<|\mathcal{P}(S)|</math>, where <math>\mathcal{P}(S)</math> is the [[power set]] of <math>S</math>. First, we note that the [[Cantor set]] <math>\mathcal{C}</math> has cardinality <math>2^{\aleph_{0}}>\aleph_{0}</math>, and since <math>\mathcal{C}\subset\mathbb{R}</math>, there is an injection <math>f:\mathcal{C}\rightarrow\mathbb{R}</math> and thus <math>|\mathbb{R}|\geq 2^{\aleph_{0}}>\aleph_{0}</math>, so <math>\mathbb{R}</math> is uncountable. In fact, it turns out that <math>|\mathbb{R}|=2^{\aleph_{0}}</math>. |
==See Also== | ==See Also== | ||
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* [[Infinite]] | * [[Infinite]] | ||
* [[Finite]] | * [[Finite]] | ||
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| + | {{stub}} | ||
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| + | [[Category:Definition]] | ||
| + | [[Category:Set theory]] | ||
Latest revision as of 19:53, 13 October 2019
A set 
is said to be uncountable if there is no injection 
. Assuming the Axiom of choice, 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 
.
Proof that 
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 
is countable. Let 
be any enumeration of the elements of 
(in other words, take an injection 
, and denote 
).
Consider the decimal expansion of each 
, say 
for all 
. Now construct a real number 
, by choosing the digit 
so that it differs from 
by at least 3 and so that is never equal to 9 or 0. It follows that 
differs from by at least 
, so 
for every . Thus, 
. However, is clearly a real number between 0 and 1, a contradiction. Thus our assumption that is countable must be false, and since 
we have that is uncountable.
An alternative proof uses Cantor's Theorem, which says that for all sets , we have 
, where 
is the power set of . First, we note that the Cantor set 
has cardinality 
, and since 
, there is an injection 
and thus 
, so is uncountable. In fact, it turns out that 
.
See Also
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