Let $(X,\tau)$ be connected such that $\emptyset$ and $X$ are the only open connected subsets. Does this imply that $\tau = \{\emptyset, X\}$?

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    $\begingroup$ 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.. $\endgroup$ – Praphulla Koushik Dec 22 '18 at 15:57
  • $\begingroup$ 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. $\endgroup$ – Dominic van der Zypen Dec 22 '18 at 16:35
  • $\begingroup$ 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. $\endgroup$ – Gerhard Paseman Dec 22 '18 at 17:07
  • $\begingroup$ Maybe edit the title to prepend "Nontrivial" $\endgroup$ – Neal Dec 22 '18 at 17:26
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    $\begingroup$ 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 $\tau = \{0, X, Y_w \}$. $\endgroup$ – 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 $X_0$.


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. $\square$

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    $\begingroup$ I would like to accept both your and Nik Weaver's answer, as they provide similar and interesting arguments, but Nik's answer was first, so I accept his answer, and +1 both of your answers. I hope you understand $\endgroup$ – Dominic van der Zypen Dec 22 '18 at 17:52

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