Connected topological space $X$ such that $\emptyset, X$ are the only open connected subsets Let $(X,\tau)$ be connected such that $\emptyset$ and $X$ are the only open connected subsets. Does this imply that $\tau = \{\emptyset, X\}$?
 A: This is false.
Example. Let $Y$ be a totally disconnected topological space in which no point is open; for example $Y = \mathbb Z_p$. Construct $X$ from $Y$ by adding a closed point in the closure of every nonempty open: $X = Y \cup \{x\}$, where $U \subseteq X$ is open if and only if $U = X$ or $U \subseteq Y$ open.
Because the only open set containing $x$ is $X$, we see that every open cover of $X$ has to contain $X$ as one of its opens, so in particular $X$ is connected.
On the other hand, every strict open $U \subsetneq X$ is contained in $Y$, hence is totally disconnected. Since points are not open by assumption, $U$ is disconnected. $\square$
A: Let $X_0$ be any topological space in which no nonempty open set is connected, e.g., $X_0 = \mathbb{Q}$ with the usual topology. Then let $X = X_0 \cup \{\infty\}$ with open sets $\{$all open subsets of $X_0\} \cup \{X\}$. The whole space is connected because the only open set containing $\infty$ is the whole space, and any proper nonempty open subset cannot contain $\infty$ and hence must be disconnected by the assumption on $X_0$.
