# 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 $$\tau = \{\emptyset, X\}$$?

• I am thinking this is simple, but I could not give an Yes/No answer.. D: I am over simplifying the question I guess... :D Can you please tell me why this question is not simple.. – Praphulla Koushik Dec 22 '18 at 15:57
• I can't say for sure that this is non-trivial... It has happened to me that I have asked utterly trivial questions on MO and had to remove them. Hopefully, this question here is somewhat interesting. – Dominic van der Zypen Dec 22 '18 at 16:35
• If U is a proper and nonempty subset of X which is open, then the complement of U in X is not open, which suggests the topology may be an ultrafilter or a subset of one. Have you considered a non principal ultrafilter as a candidate for a topology? Gerhard "It's One Or The Other" Paseman, 2018.12.22. – Gerhard Paseman Dec 22 '18 at 17:07
• Maybe edit the title to prepend "Nontrivial" – Neal Dec 22 '18 at 17:26
• At this point, the question has been completely answered. I just want to highlight two simple facts. (1) If $X$ is a finite topological set, then any non-empty open subset of minimal cardinality is connected as it cannot split into smaller non-empty open subsets. Thus the implication holds. (2) We can build for any infinite set $X$ a non-trivial countable topology $\tau$ with the desired property. Indeed, take $Y$ an infinite proper subset of $X$ and write $Y = Y_0 \cup Y_1$ where the $Y_i$ are disjoint and infinite. Then iterate by splitting each set $Y_i$ to get $\tau = \{0, X, Y_w \}$. – Luc Guyot Dec 22 '18 at 20:12

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 $$X_0$$.
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. $$\square$$