Let $X$ be a topological space and $X=X_1 \cup X_2$ with $X_1, X_2$ nonempty open irreducible subsets. Then $X$ is irreducible iff $X_1 \cap X_2 \ne \emptyset$.

The easy part: if it were $X_1 \cap X_2 = \emptyset$ then we would have $$ X = (X \setminus X_1) \cup (X \setminus X_2) $$ and this is impossible since $X$ is irreducible, so it can't be written as a union of two proper closed subsets.

The otherway gives me some problems. Suppose by contradiction $X=C_1 \cup C_2$ with $C_i$ proper closed subsets. Then... what can I do?

Could you please provide any hints, please? Thanks.


HINT: If $X$ is not irreducible, there are non-empty open sets $U$ and $V$ such that $U\cap V=\varnothing$. Clearly $(U\cap X_i)\cap(V\cap X_i)=\varnothing$ for $i=1,2$, so without loss of generality $U\cap X_2=\varnothing=V\cap X_1$. (Why?) Now consider the open sets $U=U\cap X_1$ and $X_1\cap X_2$.

  • $\begingroup$ Thank you very much. The first part and the conclusion are extremely clear. Indeed, consider the sets $U$ and $X_1 \cap X_2$ (which are open in $X_1$) and write $X_1 = (X_1 \setminus U) \cup (X_1 \setminus (X_1 \cap X_2))$: this is absurd since $X_1$ is irreducible. The only thing I can't get is how to answer to your question "why?". Indeed, I thought about it but I don't manage to prove it. Why can we suppose $U \cap X_2 = V \cap X_1 = \emptyset$? Thanks. $\endgroup$ – Romeo Jul 8 '13 at 22:11
  • $\begingroup$ @Romeo: You’re very welcome. $U\cap X_1$ and $V\cap X_1$ cannot both be non-empty, since $X_1$ is irreducible; without loss of generality assume that $V\cap X_1=\varnothing$. Then $V\subseteq X_2$, and $X_2$ is irreducible, so $U\cap X_2$ must be empty. $\endgroup$ – Brian M. Scott Jul 8 '13 at 22:28
  • $\begingroup$ Oh yes, you're right. Thank you again for your kind help. $\endgroup$ – Romeo Jul 9 '13 at 9:24
  • $\begingroup$ @Romeo: You're welcome. $\endgroup$ – Brian M. Scott Jul 9 '13 at 9:49

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