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If $ Y $ is irreducible set so is $cl(Y)$.

If $cl(Y)$ is reducible then $cl(Y)= A \cup B$ where both $A$ and $B$ is closed in $cl(Y)$. Now how do we proceed?

Ri-Li
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  • What do closed sets in $\overline{Y}$ look like? What do they look like in $Y$? – D_S May 26 '15 at 21:01

2 Answers2

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Suppose $Y$ is an irreducible subset of a larger topological space $X$. If $\overline{Y} = A \cup B$, with $A$ and $B$ closed sets of $\overline{Y}$, then $A$ and $B$ are actually closed sets in $X$. Then $$Y = Y \cap \overline{Y} = Y \cap (A \cup B) = (Y \cap A) \cup (Y \cap B)$$ with $Y \cap A$ and $Y \cap B$ closed subsets of $Y$. Since $Y$ is irreducible, one of these closed sets, say $Y \cap A$, is equal to $Y$. Thus $Y \subseteq A$, and hence $\overline{Y} \subseteq A$. But then $\overline{Y} = A$.

D_S
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Suppose $cl(Y)$ is reducible so that $cl(Y)=A\cup B$. Then $Y\subset A\cup B$ so $Y\subset A$ or $Y\subset B$. Assume the former. But $A$ is closed so $cl(Y)\subset A$ and $A\subset cl(Y)$ so $A=cl(Y)$.

Eoin
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