Consider the space $(\Bbb R, \tau_l)$, where $\tau_l$ is the lower limit topology. Is this space locally connected?
I first thought that this would be true since $\tau_l$ is finer than the standard topology, and if we pick $x \in \Bbb R$ and a neighborhood $[a,b)$ of $X$, then $x \in (a,b) \subset [a,b)$ and $(a,b)$ would be connected.
I guess this is wrong since there is a result which states that $X$ is locally connected if and only if the components of every open set of $X$ are open.
According to this since the components of $(\Bbb R, \tau_l)$ are all singletons and $\{a\}^c = (-\infty, a) \cup (a, \infty)$ is open $\{a\}$ must be closed so none of the components are open and thus the space isn't locally connected.
Why is the first approach false? Is it since that $(a,b)$ isn't connected in this topology? If so why isn't it not?
Yes, $(a,b)$ is not connected in this topology. Take $c\in(a,b)$. Then $[c,b)$ is a subset of $(a,b)$ which is both open and closed, but which is neither $\emptyset$ nor $(a,b)$.