(A question regarding:) the graph associated with an open cover of a topological space. Let $X$ denote a topological space and suppose that $\mathcal{O}$ is an open cover of $X$. Assume $\emptyset \notin \mathcal{O}.$ (Thanks Niels!) Now make $\mathcal{O}$ into an (undirected) graph as follows:


*

*Vertexes: elements of $\mathcal{O}$.

*Edges: we draw an arc between two vertexes iff they have a non-empty intersection when viewed as subsets of $X$.



Question. Is it true that if $X$ is connected (as a topological space), then $\mathcal{O}$ is connected (as a graph)?

The case where the open cover $\mathcal{O}$ has only two elements is easy, but the general case where $\mathcal{O}$ is allowed to have arbitrary cardinality is harder.
 A: Yes. Say $E_0\in\mathcal O$. Let $\mathcal O_0$ denote the set of $E\in\mathcal O$ such that $E$ is connected to $E_0$ by a chain of vertices in that graph. Let $\mathcal O_1=\mathcal O\setminus \mathcal O_0$. Let $V_j$ be the union of the $E\in \mathcal O_j$, $j=0,1$. So the $V_j$ are open and $X=V_0\cup V_1$.
Now if $E\in\mathcal O_0$, $F\in\mathcal O$ and $E\cap F\ne\emptyset$ then the definition of $\mathcal O_0$ shows that $F\in\mathcal O_0$. Which is to say that no element of $\mathcal O_0$ can intersect any element of $\mathcal O_1$. Which says that $V_0\cap V_1=\emptyset$.
Since $X$ is connected and $V_0\ne\emptyset$ this shows that $V_1=\emptyset$ and hence $\mathcal O_1=\emptyset$; so $\mathcal O=\mathcal O_0$, saying the graph is connected.
A: This seems to be true.
Note that if $K \subset \mathcal{O}$ then $\bigcup K$ is an open subset
of $X$. Also, if $K_1$ and $K_2$ are (the vertex sets of) distinct components
 of $\mathcal{O}$,
then $\bigcup K_1 \cap \bigcup K_2 = \emptyset$. Thus the decomposition
of $\mathcal{O}$ into components induces a decomposition of $X$ into
open sets.
