Is every open interval a union of half open intervals? I am reading lower limit topology on Wikipedia, which states that the lower limit topology 

[...] is the topology generated by the basis of all half-open intervals $[a,b)$, where a and b are real numbers. [...] The lower limit topology is finer (has more open sets) than the standard topology on the real numbers (which is generated by the open intervals). The reason is that every open interval can be written as a (countably infinite) union of half-open intervals.

I cannot see how to write $(a,b)$ as a countably infinite union of half-open intervals.
 A: $$(a,b) = \bigcup \{[x,b): a < x < b \}$$
Every $[x,b) \subseteq (a,b)$ whenever $a < x < b$ for the right to left inclusion, and on the other hand, if $a < x < b$, $x \in [x,b)$, which shows the left to right inclusion. If you want a countable union at all cost (topologies are closed under all unions, but maybe you're doing measure theory?) then take $x$ to be all rationals in $(a,b)$ so 
$$(a,b) = \bigcup \{[q,b): q \in \mathbb{Q} \text{ and } a < q < b\}$$ 
but then the proof of equality is a little more involved: the right to left inclusion stays the same, but if $p \in (a,b)$ we first pick $q\ in \mathbb{Q}$ with $a < q < p$, and note that $p \in [q,b)$ which is a subset of the right hand side.
A: One can write 
$$
(a,b) = \bigcup_{n\geq 1}\left[a+\frac{b-a}{2n},b\right)
$$
One inclusion is straighforward. For the non trivial one, pick $t \in (a,b)$, and $n$ large enough such that $t > a +\frac{b-a}{2n}$, so that $t \in \left[a+ \frac{b-a}{2n},b\right)$.
A: Here's an example of a union of disjoint intervals:
$$ (a,b) = \bigcup_{n=1}^\infty\left[a+\frac{b-a}{n+1}, a+\frac{b-a}n\right) $$
I wonder, why nobody presented it yet?
A: If $M = (a,b)\cap \mathbb{Q}$ then $$(a,b) = \bigcup _{c\in M} [c,b)$$
A: $$(a,b) = \bigcup_{n=1}^\infty \, \left[\, \left(1-\frac{1}{n}\right)\, a + \frac{1}{n} b   ,\, b\right)
$$
