Let $X' \subseteq X$ be a subspace of a topological space $X.$ Then the following are equivalent.
$(1)$ $X'$ is irreducible.
$(2)$ For all open sets $U,V \subseteq X$ with $U \cap X', V \cap X' \neq \varnothing$ such that $U \cap V \cap X' \neq \varnothing.$
$(3)$ The closure $\overline {X'}$ is irreducible.
I have proved $(1) \implies (2)$ from the definition of irreducible topological space and also proved $(1) \implies (3)$ by using the fact that any non-empty open set in a irreducible topological space is dense. How can I prove $(2) \implies (3)$ and $(3) \implies (1)$?
Any help will be highly appreciated. Thank you very much.
Best Answer
(2) as stated is a wrong equivalence: it holds for all spaces $X$ and every non-empty subset $X'\subseteq X$: just take $U=V=X$ and it's trivially satisfied.
You probably mean
This is clear because we can restate it (by the definition of the subspace topology )
And this is a dual statement to $X'$ is irreducible in the subspace topology. Just assume disjoint open non-empty sets exist; then their complements are proper closed subsets that cover $X'$, and vice versa.
Suppose (2) holds in the correct version as above. We want to see $\overline{X'}$ is irreducible so assume $A \cup B = \overline{X'}$ with $A,B$ closed (in $X$ or $\overline{X'}$, it's equivalent) Then if $O_A=\overline{X'} \setminus A$ which is open in $\overline{X'}$ were non-empty, it would intersect $X$ (which is dense in $\overline{X'}$) in a non-empty relatively open set, and likewise for $O_B=\overline{X'}\setminus B$. But they cannot both intersect $X$ non-emptily because they are disjoint and this would contradict (2). So one of $O_A$ or $O_B$ is empty, and thus equals $\overline{X'}$, as required.
(3) to (1): suppose $A \cup B= X'$ where $A,B$ are relatively closed. It follows (check this) that $\overline{A} \cup \overline{B} = \overline{X'}$ (closures taken in $\overline{X'}$) and so one of the closures must equal $\overline{X'}$ by (3), and that set must thus have equaled $X'$.