Is $\mathbb R$ endowed with the left-hand topology (also called lower limit topology) path-connected?
Intuitively, I know that the answer is yes but I'm not sure how to prove it.
Would it suffice to say that a basic open set in $\mathbb R_\ell$ is of the form $[a,b)\in \mathbb R$ and is homeomorphic to I=[0,1]? Therefore any open set in $\mathbb R_\ell$ is path-connected?
Note that $\mathbb{R}$ with the lower-limit topology is not connected, and even satisfies the stronger "disconnectivity" property of zero dimensionality, meaning that it has a base consisting of clopen (closed and open) sets. For $a <b$ we have $$\mathbb{R} \setminus [a, b) = ( - \infty, a) \cup [b,\infty)$$ which can easily be seen to be open in this topology.
Because it is not connected, it cannot be path-connected, as if there is a path between two points, those two points must belong to the same connected component of the space.