# Connected topologies on $\mathbb{R}$ strictly between the usual topology and the lower-limit topology

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.

### Definitions

• 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 $$\{ [a,b) : a,b \in \mathbb{R} , a < b \}$$.

• 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. – freakish Dec 6 '18 at 9:38
• @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. – stochastic randomness Dec 6 '18 at 9:42
• Ah yes, you're right. – 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 $$(\Bbb R,\sigma)$$.