Can the Euclidean unit interval have a finer connected topology?

Question

Consider the unit interval $X = [0,1]$ equipped with the Euclidean topology $\tau_E$, and consider some other topology $\tau_F$ on $X$ that is strictly finer than $\tau_E$.

Does there exist such a topology $\tau_F$ for which $(X,\tau_F)$ is a connected space?

I suspect there is either 1) a simple, abstract way of showing this topology cannot exist, or 2) a somewhat complicated example showing it can exist.

PARTIAL RESULTS:

Consider a process by which (at least) one subset of $X$ is added to the Euclidean topology to form $\tau_F$. No such subset could be a finite subset of $X$, as finite subsets of $X$ are closed with respect to the Euclidean topology and therefore clopen in $(X,\tau_F)$, disconnecting the space. Therefore, $\tau_F$ cannot contain any finite subsets of $X$.

Now consider adding (at least) one subset interval of $X$ that is not in $\tau_E$ to $\tau_E$ to form $\tau_F$. Such a subset cannot be an open interval, as this is already included in $\tau_E$. Such a subset cannot be a closed interval, and the resulting set would be clopen in $\tau_F$, disconnecting the space. Lastly, it cannot be a half-open interval $[a,b)$ or $(a,b]$, as the space can always be split into two disjoint open sets of $\tau_F$ of the following form:

$$X = [0,a)\cup\left([a,b)\cup(\frac{a+b}{2},1]\right) = [0,a)\cup[a,1]$$

or

$$X = \left([0,\frac{a+b}{2})\cup(a,b]\right)\cup(b,1] = [0,b]\cup(b,1]$$

As a result, $\tau_F$ must also not contain any half-open/half-closed or closed intervals.

Answer 1

I believe the following is a connected topology on the unit interval that is finer than the Euclidean topology.

Define $\tau_F$ as the topology created by a topological basis consisting of 1) every open ball in $\tau_E$ as well as 2) every subset of rational numbers within each open ball in $\tau_E$. Each of the subsets described by 2) is neither an open or closed set in $\tau_E$, and their union with any other subset in the basis is similarly neither open or closed in $\tau_E$.

The closed sets of the topological space $(X,\tau_F)$ must therefore always be either closed intervals or intersections of closed intervals with sets of irrational numbers contained within some closed interval. All of these must contain at least one irrational number; since this is the case, no closed set of $(X,\tau_F)$ can simultaneously belong to the "new" family of open sets defined by 2).

As a result, the only clopen sets in $\tau_F$ are $\varnothing$ and $X$, and so the space is connected.

Answer 2

Here’s another counterexample — a concretely described topology finer than the Euclidean one but not connected.

We’ll add one new basic open set, $\newcommand{\f}[1]{\frac{1}{#1}} Z = \{0\} \cup \left( \bigcup_{k=1}^\infty (\f{2k+1},\f{2k}) \right)\cup \left( \bigcup_{k=1}^\infty (-\f{2k},-\f{2k+1})\right) $. So $Z$ is open everywhere except $0$, and contains $0$, but every neighbourhood of $0$ meets both $Z$ and $\mathbb{R} \setminus Z$. Now take $\newcommand{\T}{\mathcal{T}}\T_Z$ to be the topology consisting of all sets of the form $U_1 \cup (Z \cap U_2)$, where $U_1, U_2$ are open in the usual Euclidean topology $\T_E$. (Exercises: show this is indeed a topology; show that any topology extending $\T_E$ and containing $Z$ must extend $\T_Z$; conclude that $\T_Z$ is the least topology extending $\T_E$ and containing $Z$.)

Note that if $U$ is open in $\T_Z$, then $U \cap (-\infty,0)$ and $U \cap (0,\infty)$ are both open in the usual sense (intuitively, away from $0$ the topology is unchanged, $Z$ was already open there); and if moreover $0 \in U$, then $Z \cap (-\varepsilon,\varepsilon) \subseteq U$ for some $\epsilon$, so in particular $U$ meets both $(\infty,0)$ and $(0,\infty)$.

Now we can see that $\T_Z$ is connected. It suffices to show that any clopen set containing $0$ is all of $\mathbb{R}$, since then for any clopen, either it or its complement is $\mathbb{R}$. So suppose $U$ is clopen and contains $0$; we know $U$ meets both $(-\infty,0)$ and $(0,\infty)$. So $U \cap (0,\infty)$ is clopen and non-empty in $(0,\infty)$, so is all of $(0,\infty)$ since $(0,\infty)$ is connected; and similarly, $U \cap (-\infty,0) = (-\infty,0)$. Thus $U$ contains $(-\infty,0)$, $0$, and $(0,\infty)$, so it must be all of $\mathbb{R}$.