This is the question:
Which of these sets are open sets on the lower limit topology on $\mathbb{R}$, whose basis elements are $[a,b),a<b$?
$$[4,5)\qquad\left\lbrace3\right\rbrace\qquad [1,2]\qquad(7,8)$$
The first one clearly is open, as it belongs to the basis.
The secon one can't be, as all basis elements have more than one element, and so no union of basis elements can have only one element.
The fourth can be built by:
$$(7,8)=\bigcup_{7<a<8}[a,8)$$
The third one, I can't prove it's not (it's clearly not open). I'm trying something like this:
If it was open, it would be the result of union of sets like $[1,a)$. I can't prove though, that the union of those sets can't have a closed limit.
EDIT: Could it be: The union of those sets is defined by setting a set of values for $a$. $a\in\mathbb{R}$, so there must be a maximum $a$: $a_{max}$, and so the final result of the union would be $[1,a_{max})$. That works if the set of values for $a$ is finite. If it's not, then if it's bounded the same thing works, if its unbounded the set would be $[1,\infty)$
Best Answer
Note that the union of intervals of the form $[a,b)$ is never $[x,y]$. To see this, first note that it is sufficient to prove this fact for increasing unions of $[a,b_i)$.
Suppose now that $b_i$ is a strictly increasing sequence of real numbers, and $a<b_0$, then $(a,b_i)$ are open sets in the Euclidean topology, and so $\bigcup(a,b_i)=(a,\sup b_i)$ and it is easy to see that $\bigcup[a,b_i)=\{a\}\cup\bigcup(a,b_i)$.
Similarly $\{3\}$ is not open either, because it is the interval $[3,3]$.