Finding the closure of sets in different topologies. $\newcommand{\N}{\mathbb{N}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\T}{\mathcal{T}}
\newcommand{\R}{\mathbb{R}}$
Problem 
Let $\T_1$ be the lower limit topology in $\R$ and $\T_{2}$ be the topology generated by basis $\mathcal{C}=\{[a,b)|a<b,a,b\in\Q\}$. Determine the closures of the intervals $A=(0,\sqrt{2})$ and $B=(\sqrt{2},3)$ in these two topologies. 
Attempted Solution
（I shall use $A'$ to denote the set of limit points of set $A$）
In $\T_{1}$, we claim that $\overline{\left(0,\sqrt{2}\right)}=[0,\sqrt{2})$.
To show this, we first show that $\left\{ 0\right\} $ is a limit
point of $\left(0,\sqrt{2}\right).$ Let $U$ be any open set containing
in $\left\{ 0\right\} $, then $\exists[a,b)$ such that $\left\{ 0\right\} \in[a,b)\subset U.$
Therefore, $\exists\varepsilon>0$ such that $[0,\varepsilon)\in[a,b),$
which implies $U\cap\left(0,\sqrt{2}\right)\neq\emptyset$. Since
$U$ is arbitrary, we have that $\left\{ 0\right\} \in\left(0,2\right)'$.
Then we have $[0,\sqrt{2})\subset\overline{\left(0,\sqrt{2}\right)}.$
To see the other direction, it suffices to show that $[0,\sqrt{2})$
is closed. Indeed, $[0,\sqrt{2})^{c}=\left(-\infty,0\right)\cup[\sqrt{2},+\infty)=\cup_{n=1}^{\infty}\left[[-n,0)\cup[\sqrt{2},n)\right]$
which is open and thus $[0,\sqrt{2})$ is closed and contains $\left(0,\sqrt{2}\right)$.
Then by definition, $\overline{\left(0,\sqrt{2}\right)}\subset[0,\sqrt{2})$.
By virtually the same argument, we can show that $\overline{\left(\sqrt{2},3\right)}=[\sqrt{2},3)$.
In $\T_{2}$, we claim that $\overline{\left(0,\sqrt{2}\right)}=\left[0,\sqrt{2}\right].$
Using a similar argument, we can show that $\left\{ 0\right\} \in\left(0,\sqrt{2}\right)'$.
To show that $\left\{ \sqrt{2}\right\} \in\left(0,\sqrt{2}\right)'$
, we note that for any open set $U$ containing $\left\{ \sqrt{2}\right\} $,
$\exists[a,b)\in U$ such that $a<b$ and $a,b\in\Q.$ This guarantees
that $a<\sqrt{2}$ since $\sqrt{2}\notin\Q$. Thus $U\cap\left(0,\sqrt{2}\right)\neq\emptyset.$
Since $U$ is arbitrary, $\sqrt{2}\in\left(0,\sqrt{2}\right)'$. We
then have $\left[0,\sqrt{2}\right]\in\overline{\left(0,\sqrt{2}\right)}.$
To see this other direction, it suffices to show that $\left[0,\sqrt{2}\right]$
is closed. To see this, we note that we can rewrite $[0,\sqrt{2}]^{c}=\cup_{n=1}^{\infty}\left[[-n,0)\cup[a_{n},+\infty)\right]$
where $a_{n}\in\Q$ for all $n\in\N,$ and $a_{n}\downarrow\sqrt{2}.$
Notice that the latter is open as it's a countable union of open sets
in $\T_{2}$, whence $[0,\sqrt{2}]$ is closed, and as a result $\overline{\left(0,\sqrt{2}\right)}\subset[0,\sqrt{2}]$.
By a similar argument, we can show that $\overline{\left(\sqrt{2},3\right)}=[\sqrt{2},3)$ 
Question
I don't know if my arguments are convincing enough. I hope someone can take a look at my attempt and point out any mistakes and possible improvements. 
 A: $\newcommand{\N}{\mathbb{N}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\T}{\mathcal{T}}
\newcommand{\R}{\mathbb{R}}$The definitions of $\mathcal{T}_1$ and $\mathcal{T}_2$ are similar.  The generated by intervals of the real line:


*

*$\mathcal{T}_1$ generated by half-open intervals $\{[a,b)|\,a<b  \text{ with }a,b\in\R\}$

*$\mathcal{T}_2$ generated by half-open intervals $\{[a,b)|\,a<b \text{ with }a,b\in\Q\}$
So what difference does it make if we only use rational numbers $\mathbb{Q}$ instead of every possible real number $\mathbb{R}$?
In this system, can the closures contain the left points?  $0 \in \overline{(0, \sqrt{2})}$ or $\sqrt{2} \in \overline{(\sqrt{2}, 3)}$ ?


*

*if $0 \in [a,b)$ then $ a \leq 0 < b$, so that $[a,b)$ has some positive numbers

*if $\sqrt{2} \in [a,b)$ then $ a \leq \sqrt{2} < b$ so the interval $[a,b)$ can't avoid sharing some points in $(\sqrt{2},3)$
Both cases had the same outcome, regardless if we used $\mathbb{R}$ or $\mathbb{Q}$... What about the points on the right?


*

*if $\sqrt{2} \in [a,b)$ then $ a \leq \sqrt{2} < b$ but if $a \in \mathbb{Q}$ then $a < \sqrt{2}$ since $\sqrt{2}$ is not a fraction.


*

*$(0, \sqrt{2}) \cap [\sqrt{2}, \infty) = \varnothing $

*This is acceptable in $\mathcal{T}_1$ but not $\mathcal{T}_2$.



Let's draw the perfect non-overlap in this case, that both $\mathcal{T}_1$ always permits but $\mathcal{T}_2$ only sometimes:
[oooooooooooooooooooo)................
.....................[ooooooooooooooo)

vs                   3?

(oooooooooooooooooooo)................
.....................(ooooooooooooooo)    

