Are singletons closed or open in the lower limit topology? Recall $(\mathbb{R}, \mathcal{T}_\text{lower limit})$ where lower limit topology $\mathcal{T}_\text{lower limit} = \mathcal{T_\mathcal{B}}$ where
$\mathcal{B} = \{[a,b) \subseteq \mathbb{R}, a < b\}$

Question: Are singleton sets $\{a\}, a \in \mathbb{R}$ open?

Attempt: 

$\{a\}^c = \mathbb{R}\backslash\{a\} = (-\infty, a) \cup (a, \infty)
 \in \mathcal{T}_{lowerlimit}$
So $\{a\}$ is closed

Correct?
 A: It's true that singletons are closed. But singletons could also be open. You haven't shown. It is true that if all singletons are open, the space is discrete, which it is not. But there still could be some freak singleton that is open, maybe? Showing something is closed does not mean it's not open: in the lower limit topology $[0,1)$ is both open and closed! So try a direct proof, not using closedness:
Suppose $\{p\}$ is open. Then there must be a basic open set of the lower limit topology, i.e. a set of the form $[a,b)$, such that $p \in [a,b) \subseteq \{p\}$. In other words, all points in this $[a,b)$ must be equal to $p$. But this is clearly absurd as $[a,b)$ is infinite (or just notice that both $a$ and $a + \frac{b-a}{2}$ are in it). So $\{p\}$ cannot be open. And this holds for any $p \in \mathbb{R}$.
A: You are correct that singletons are closed, but it might be worthwhile to show why exactly $(-\infty, a)$ and $(a, \infty)$ are open for each $a \in \mathbb{R}$.  Use sets of the form $[-n, a)$ for $(-\infty, a)$ and something similar for the other.  Note that singletons can't be open in this topology.  If they were, then what could you say about any arbitrary subset $A \subseteq \mathbb{R}$?
