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<record version="10" id="1109">
 <title>irreducible</title>
 <name>IrreducibleClosedSet</name>
 <created>2001-12-20 01:41:56</created>
 <modified>2006-02-17 11:23:10</modified>
 <type>Definition</type>
 <creator id="2727" name="mathcam"/>
 <author id="2727" name="mathcam"/>
 <author id="1858" name="matte"/>
 <author id="3" name="drini"/>
 <author id="62" name="nerdy2"/>
 <classification>
	<category scheme="msc" code="54B05"/>
	<category scheme="msc" code="14A10"/>
	<category scheme="msc" code="14A15"/>
 </classification>
 <defines>
	<concept>reducible</concept>
 </defines>
 <related>
	<object name="IrreducibleComponent"/>
	<object name="HyperconnectedSpace"/>
 </related>
 <keywords>
	<term>topological space</term>
 </keywords>
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 <content>A subset $F$ of a topological space $X$ is {\em reducible} if it can be written as a union $F = F_1 \cup F_2$ of two closed proper subsets $F_1$, $F_2$ of $F$ (closed in the subspace topology).  That is, $F$ is reducible if it can be written as a union $F = (G_1\cap F)\cup(G_2\cap F)$ where $G_1$,$G_2$ are closed subsets of $X$, neither of which contains $F$.

A subset of a topological space is {\em irreducible} (or \emph{hyperconnected}) if it is not reducible.

As an example, consider $\{ (x,y)\in\mathbb{R}^2 : xy = 0 \}$ with the subspace topology from $\mathbb{R}^2$.  This space is a union of two lines $\{ (x,y)\in\mathbb{R}^2 : x = 0 \}$ and $\{ (x,y)\in\mathbb{R}^2 : y = 0 \}$, which are proper closed subsets.  So this space is reducible, and thus not irreducible.</content>
</record>
