3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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 $(\Bbb R,\sigma)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.