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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.

Steven-Owen
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3 Answers3

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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.

Alex Becker
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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
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    @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
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    @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 $A \subseteq Y \cap Y'$. –  Feb 25 '13 at 23:16
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    @user315 typo correction $A\subseteq Y\cup Y’$. – user264745 Sep 11 '24 at 07:42
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Consider $S^c$ and $T^c$, which are open in $\bar{Y}$. Then, since any nonempty open subset of an irreducible space ($Y$) is dense, we have, $\bar{S^c\bigcap Y}=Y$, and $\bar{T^c\bigcap Y}=Y$. Then, $Y=\bar{(S^c\bigcap Y})\bigcup(\bar{T^c\bigcap Y})$, which is a contradiction to Y being irreducible.

Baby elephant
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Here are multiple equivalent definition of irreducible. I will use following definition:

A topological space $X$ is said to be irreducible if every pair of non empty open sets in $X$ intersect.

Clearly, this definition is equivalent to the decomposition, just take complement in $X$. We want to show $\overline{Y}$ is irreducible. Let $\overline{Y}\cap U_i\neq \emptyset$ be open set in subspace topology $\overline{Y}$, $i=1,2$. By equivalent definition of closure, $Y\cap U_i\neq \emptyset$. Since $Y$ is irreducible, we have $Y\cap U_i$’s intersect. Thus $\overline{Y}\cap U_i$’s intersect because $Y\cap U_i\subseteq \overline{Y}\cap U_i$. Hence $\overline{Y}$ is irreducible.

user264745
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