I was wondering if the following is true. If I have two closed sets $A$ and $B$ such that $A$ is irreducible. Say we take $C = cl(A - B)$ is it true that because $A$ is irreducible $C = A$?
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What if $A = B$? – Rob Arthan May 03 '22 at 22:14
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What is the definition of an $irreducible$ closed set? – DanielWainfleet May 03 '22 at 22:20
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1@DanielWainfleet: see https://math.stackexchange.com/questions/526147/irreducible-subsets-of-a-topological-space – Rob Arthan May 03 '22 at 22:23
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This is the fact that nonempty open sets in irreducible spaces are dense. In the notation of the question, we have $$A\supseteq C\cup (B\cap A) \supseteq (A-B)\cup (B\cap A) = A$$ and thus $C\cup (B\cap A) = A$. Now, both $C$ and $B\cap A$ are closed in $A$ while $A$ is irreducible, so $$C = A\quad\text{or}\quad B \supseteq A$$
Karl Kroningfeld
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