# Equivalent definition of irreducible topological subspace.

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.

(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 $$X'$$.

• 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? – Dbchatto67 Feb 21 at 3:28
• @Dbchatto67 which $A$ is closed in $X’$? – Henno Brandsma Feb 21 at 5:03
• 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. – Dbchatto67 Feb 21 at 5:51
• 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 $X'.$ – Dbchatto67 Feb 21 at 6:02
• @Dbchatto67 irreducible works with relatively closed subsets. – Henno Brandsma Feb 21 at 6:08