# Show that $X$ is a irreducible topological space.

A topological space $$X$$ is said to be irreducible if for any decomposition of $$X$$ with $$X=A \cup B$$ where $$A,B$$ are closed subsets of $$X$$ we either have $$X=A$$ or $$X=B.$$

Theorem $$:$$

Let $$X$$ be a topological space such that any non-empty open subset of $$X$$ is dense in $$X.$$ Then show that $$X$$ is irreducible.

My attempt $$:$$

Suppose $$X=A \cup B$$ with $$A$$ and $$B$$ closed in $$X.$$ If either $$X=A$$ or $$X=B$$ we are through. So WLOG let us assume that $$A \subsetneq X$$ and $$B \subsetneq X.$$ Then $$X \setminus A$$ and $$X \setminus B$$ are non-empty open subsets of $$X.$$ So by the given hypothesis they are both dense in $$X.$$ Therefore \begin{align} \overline {X \setminus A} & = \overline {X \setminus B} = X. \\ \implies X \setminus \operatorname {int} (A) & = X \setminus \operatorname {int} (B) = X. \\ \implies \operatorname {int} (A) & = \operatorname {int}(B) = \emptyset. \end{align}

So $$A$$ and $$B$$ are nowhere dense subsets of $$X.$$ Therefore $$X$$ is meagre or a set of first category.

Now I got stuck at this stage. Would anybody please help me to proceed further?

Thank you very much.

• Inspect int(A union B) – Rob Bland Feb 19 at 17:11
• I think $A \cup B$ is also nowhere dense and therefore $\operatorname {int} (A \cup B) = \emptyset.$ But then $X = \emptyset.$ – Dbchatto67 Feb 19 at 17:17
• You are right- the union of two nowhere dense sets is nowhere dense; a more general version of this statement is called the Baire Category theorem, & the proof of this simple case is easy to do by taking complements & proving the intersection of open-dense sets is also open-dense – Rob Bland Feb 19 at 17:20

As $$X=A \cup B$$ and assume wlog $$A\not =X$$. We know that $$X\setminus A\subseteq B$$. $$X\setminus A$$ being dense implies that $$X$$ is the smallest closed set containing $$X\setminus A$$. Nevertheless, $$B$$ is a closed which contains $$X\setminus A$$, too. Hence, we get $$X=B$$ what we wanted to show.
The union of two nowhere dense sets is nowhere dense and $$X$$ clearly is not nowhere dense $$\operatorname{int}(X)=X \neq \emptyset$$). So already have your contradiction! You just don't realise it yet.