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<record version="8" id="8066">
 <title>numerable set</title>
 <name>NumerableSet</name>
 <created>2006-06-20 15:05:21</created>
 <modified>2009-09-28 20:35:00</modified>
 <type>Definition</type>
 <creator id="12619" name="juanman"/>
 <author id="3771" name="CWoo"/>
 <author id="12619" name="juanman"/>
 <classification>
	<category scheme="msc" code="97A80"/>
 </classification>
 <defines>
	<concept>enumeration</concept>
	<concept>enumerable</concept>
 </defines>
 <related>
	<object name="Calculus"/>
	<object name="TopicsOnCalculus"/>
	<object name="Denumerable"/>
	<object name="Countable"/>
 </related>
 <keywords>
	<term>Analysis</term>
 </keywords>
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 <content>Let $X$ be a set.  An \emph{enumeration} on $X$ is a surjection from the set of natural numbers $\mathbb{N}$ to $X$.

A set $X$ is called \emph{numerable} if there is a bijective enumeration on $X$.

It is easy to show that $\mathbb{Z}$ and $\mathbb{Q}$ are numerable.

It is a standard fact that $\mathbb{R}$ is not numerable. For, if we suppose that the numbers [0,1] were countable, we can arrange them in a list (given by the supposed bijection).

Representing them in a binary form, it is not hard to construct an element in [0,1], which is not in the list.

This contradiction implies that [0,1]$\subset\mathbb{R}$ is not numerable. 

\textbf{Remark}.  If the enumeration $\mathbb{N}\to X$ is furthermore a computable function, then we say that $X$ is \emph{enumerable}.  There exists numerable sets that are not enumerable.</content>
</record>
