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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.

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(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

(2) for all open $U,V$ of $X$ such that $U \cap X'\neq \emptyset$ and $V \cap X'\neq \emptyset$ we have $U \cap V \cap X' \neq \emptyset$.

This is clear because we can restate it (by the definition of the subspace topology )

(2') for all non-empty $U,V$ open in $X'$ we have $U \cap V\neq \emptyset$.

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 $X'$.

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  • $\begingroup$ In $(3) \implies (1)$ if suppose $\overline {X'} = \overline {A}$ how can I say that $X'=A$? If we take two sets $(0,1)$ and $[0,1]$ then they have the same closure in $\Bbb R$ with usual topology. Are they equal? $\endgroup$ – Dbchatto67 Feb 21 at 3:28
  • $\begingroup$ @Dbchatto67 which $A$ is closed in $X’$? $\endgroup$ – Henno Brandsma Feb 21 at 5:03
  • $\begingroup$ I tried to say that if $\overline A = \overline B$ we cannot say in general that $A=B.$ So what is the reasoning here you have used to claim that $\overline {X'} = \overline {A} \implies X' = A$? Thanks. $\endgroup$ – Dbchatto67 Feb 21 at 5:51
  • $\begingroup$ Well when you have taken closure of $X'$ in $X'$ then it is $X'$ itself. Then how do you apply $(c)$ to say that $X' = \overline A$ or $X'= \overline B,$ where closures are taken in $X'.$ $\endgroup$ – Dbchatto67 Feb 21 at 6:02
  • $\begingroup$ @Dbchatto67 irreducible works with relatively closed subsets. $\endgroup$ – Henno Brandsma Feb 21 at 6:08

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