It is well-known that the usual order/metric topology on $\mathbb{R}$ is connected, and the lower-limit topology is not connected (it is even totally disconnected). We also know that the lower-limit topology is strictly finer than the usual topology.

Are there connected topologies on $\mathbb{R}$ strictly between these two? (That is, is there is a connected topology on $\mathbb{R}$ which is strictly finer than the usual topology, but coarser than the lower limit topology?)

I know that given any lower-limit basic open set $[a,b)$ (for $a < b$) the topology generated by the subbase consisting of $[a,b)$ and all of the usual open sets is not connected (because $[a,+\infty) = [a,b) \cup ( \frac{a+b}{2} , + \infty )$ and $\mathbb{R} \setminus [a,+\infty) = (-\infty , a )$ are both open in this topology). But perhaps there are more complicated lower-limit-open sets that can be added to yield a connected topology.


  • A topological space $X$ is connected if the only subsets of $X$ that are clopen (closed and open) are $\emptyset$ and $X$.

  • The lower-limit topology on $\mathbb{R}$ is the topology generated by the base $\{ [a,b) : a,b \in \mathbb{R} , a < b \}$.

  • $\begingroup$ You are right that it doesn't work for $[a,b)$. But what if you add $U=(-\infty, -1)\cup[0,\infty)$ to the standard topology? I'm writing this is as a comment because I'm not 100% sure if it works but it looks so. $\endgroup$ – freakish Dec 6 '18 at 9:38
  • $\begingroup$ @freakish Essentially the same problem. If your $U$ is open in the new topology, then so is $U \cap ( \frac{-1}{2} , +\infty ) = [0,+\infty)$, and clearly $\mathbb{R} \setminus [0,+\infty) = ( - \infty , 0 )$ is also open. $\endgroup$ – stochastic randomness Dec 6 '18 at 9:42
  • $\begingroup$ Ah yes, you're right. $\endgroup$ – freakish Dec 6 '18 at 9:45

Are there connected topologies on $\mathbb{R}$ strictly between these two?

Yes. For instance, let $\sigma$ be a topology on $\Bbb R$ generated by its standard topology $\tau$ and a set $S=\Bbb R\setminus\{-\frac 1n:n\in\Bbb N\}$. The space $(\Bbb R,\sigma)$ is connected because $\operatorname{int}_\tau A=\operatorname{int}_\sigma A$ for each closed subset $A$ of $(\Bbb R,\sigma)$.

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