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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.

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  • $\begingroup$ Inspect int(A union B) $\endgroup$ – Rob Bland Feb 19 at 17:11
  • $\begingroup$ I think $A \cup B$ is also nowhere dense and therefore $\operatorname {int} (A \cup B) = \emptyset.$ But then $X = \emptyset.$ $\endgroup$ – Dbchatto67 Feb 19 at 17:17
  • $\begingroup$ 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 $\endgroup$ – Rob Bland Feb 19 at 17:20
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I do not know whether your ansatz leads somewhere but this is the way that I would have taken.

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.

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  • $\begingroup$ Good point @Jonas Lenz. $\endgroup$ – Dbchatto67 Feb 19 at 17:22
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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.

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  • $\begingroup$ I have already realised that before your answer was posted. See my comment above. $\endgroup$ – Dbchatto67 Feb 28 at 18:40

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