A topological space is called connected if any presentation of $X$ as $X = V_1 \uplus V_2$ by disjoint open subsets implies that one of them is trivial ($V_1 = X$ or $V_2 = X$). By taking complement one can replace the word "open" by "close".
A topological space is called irreducible if any presentation of $X$ as $X = X_1 \cup X_2$ by two closed subsets implies that one of them is trivial ($X_1 = X$ or $X_2 = X$).
Clearly every irreducible space is connected. The converse is not always true but:
Proposition:
Let $X$ be a connected topological space which has an open covering by irreducible subspaces. Then $X$ is irreducible.
I want to prove this proposition.
Best Answer
HINT: It’s easily checked that a space $X$ is irreducible if and only if $U\cap V\ne\varnothing$ whenever $U$ and $V$ are non-empty open subsets of $X$. Let $X$ be connected, and let $\mathscr{U}$ be an open cover of $X$ by irreducible subspaces. Suppose that $V$ and $W$ are disjoint open subsets of $X$; you want to show that one of them is empty.
Let
$$\begin{align*} &\mathscr{U}_V=\{U\in\mathscr{U}:U\cap V\ne\varnothing\}\;,\\ &\mathscr{U}_W=\{U\in\mathscr{U}:U\cap W\ne\varnothing\}\;,\text{ and}\\ &\mathscr{U}_0=\mathscr{U}\setminus(\mathscr{U}_V\cup\mathscr{U}_W)\;. \end{align*}$$
Consider the open sets $\bigcup\mathscr{U}_V,\bigcup\mathscr{U}_W$, and $\bigcup\mathscr{U}_0$, bearing in mind that $X$ is connected.