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Start with an irreducible space $X$. Take a subset $Y$ that is irreducible. Show that the closure of $Y$ is still irreducible.

I imagine we are supposed to start with saying, assume we have a decomposition for $\bar Y = S\cup T$ and then somehow derive a contradiction to $X$ or $Y$'s irreducibility, but I am struggling.

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The irreducibility of $X$ is unnecessary. –  Alex Becker Feb 25 '13 at 22:10
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Suppose $\bar Y$ is not irreducible, so we have closed subsets $S,T$ in $X$ such that $\bar Y$ is not contained in either $S$ or $T$ but $\bar Y \subseteq S\cup T$. If $Y\subseteq S$, then $\bar Y\subseteq \bar S=S$, a contradiction, so $Y$ is not contained in $S$. Similarly $Y$ is not contained in $T$. But $Y\subseteq \bar Y\subseteq S\cup T$, thus $Y$ is not irreducible.

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That's what I originally was going for, but I thought the definition of irreducible is a decomposition into proper closed (not necessarily disjoint) subsets, not just a covering by closed sets. And from here, it is not as if we can just intersect $Y$ with $S$ and $T$, because $Y\cap S$ and $Y\cap T$ are not necessarily closed. –  Steven-Owen Feb 25 '13 at 22:24
    
@ricky They are closed in the subspace topology, which is what matters. The nature of the subspace topology allows us to work with coverings like this. –  Alex Becker Feb 25 '13 at 22:25
    
O, I see. That's confusing though going back and forth between X's topology and the induced subspace topology. Thanks for your help! –  Steven-Owen Feb 25 '13 at 22:27
    
@ricky: A subset $A \subset X$ is irreducible iff there exist subsets $Z,Z' \subset A$ _closed in the topology induced on $A$_, such that $A = Z \cup Z'$; by definition $Z$ and $Z'$ are of the form $Z = Y \cap A$ and $Z' = Y' \cap A$, respectively, for subsets $Y,Y' \subset X$ closed in $X$, hence the equality $A = (Y \cap A) \cup (Y' \cap A) = (Y \cup Y') \cap A$ which is equivalent to the inclusion $A \subseteq Y \cap Y'$. –  Adeel Feb 25 '13 at 23:16
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