# If there exists an irreducible open covering, then the space is irreducible.

If a topological space $X$ has an open covering $\{ U_i \}$ such that every $U_i$ is irreducible and that for each $i,j$, $U_i \cap U_j \neq \emptyset$, then is $X$ irreducible?

I found Space admitting an irreducible connected open covering is irreducible , but I think this is wrong.: If this answer were correct, every scheme would be affine.

• Why would every scheme then be affine? – Fredrik Meyer Apr 24 '18 at 11:50
• @FredrikMeyer He says in comment that the definition of irreducibleness is can't be union of proper open subsets. If this were true, then since every scheme admits affine open covering, the scheme must be covered by only one affine open. – k.j. Apr 24 '18 at 11:53
• @k.j. That definition is off. Irreducible spaces are the ones that are not the union of two (or, equivalently, finitely many) proper closed subsets. – user228113 Apr 24 '18 at 12:01

It is true. Let $U$, $V$ be any two non-empty open subsets of $X$. We want to show that $U\cap V\ne \emptyset$. Since $\{U_i\}$ is a covering, there must be some $i,j$ such that $U_i\cap U\ne\emptyset$ and $U_j\cap V\ne\emptyset$. Since $U_i\cap U$ and $U_i\cap U_j$ are non-empty open subsets of $U_i$ and $U_i$ is irreducible, $U_i\cap U_j\cap U=(U_i\cap U_j)\cap(U_i\cap U)\ne\emptyset$. Finally, since $U_i\cap U_j\cap U$ and $U_j\cap V$ are non-empty open subset of $U_j$ and $U_j$ is irreducible, then $U_i\cap U_j\cap V\cap U=(U_i\cap U_j\cap U)\cap(U_j\cap V)\ne\emptyset$.
I can't quite weight on the scheme part, but probably what you have for those is a situation where $U_1\cap U_2$ and $U_2\cap U_3$ are non-empty, while $U_1\cap U_3=\emptyset$.