Are these open sets?

Solution 1:

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 $[3,3]$.

Solution 2:

If the set $[1,2]$ was open, then $2$ would have an open neighborhood $[a,b)$ such that $2\in[a,b)\subset[1,2]$. Now take any element $p$ such that $2<p<b$. It belongs to $[a,b)$, hence to $[1,2]$, and is greater than $2$...

Solution 3:

If $[a, b]$ is open in the lower limit topology, it is expressible as a union of sets of the form [c, d). Clearly no $[c, d)$ where $d > b$ can be in the union or else the union would contain elements not in [a, b]. So the sets must be of the form $[c, d)$ where $d \leq b$. But then the union would not contain $b$, and so $[a, b]$ cannot be expressed as a union of basic open sets. Hence $[a, b]$ is not open.