Closure of interval in a given topology $\tau$ of $\mathbb{R}$ Given the set 
$$B=\{\emptyset\} \cup \{ [a,b):-\infty <a <b <\infty\},$$
then $B$ is a base for a topolgy $\tau$ on $\mathbb{R}$. Can someone calculate the closure of $(a,b)$ and $(a,b]$? I tried but I couldn't find the answer.
 A: We have that


*

*The closure of $(a, b)$ is $[a, b)$

*The closure of $(a, b]$ is $[a, b]$
Here's an outline on how to see this. Note first that $[a, b)$ is closed, because
$$ [a, b) = (- \infty, a)^c \cap [b, \infty)^c.$$
But $(a, b)$ is not closed (this is actually a bit tricky), whence the closure is $[a, b)$. The second point is very similar.
A: The basis you have tells you that [a,b) is an elite club, anything in the set is closer to every member of the set than anything outside the set.
This means that for a given point p, the set [p,x) is very close for any x, which means that numbers just bigger than p are closer than other numbers.
The closure of a set A is everything you can get as close as possible to and still be in A
So the closure of (a,b) is everything you can get as close as possible to and still be in (a,b). You can to everything in (a,b), and obviously you can’t get closer. You can take the numbers just above a to get into any club a is in. But you can’t get any number greater than or equal to b, and those numbers are closer to b than all the rest, so it’s not in the closure. So the closure of (a,b) is [a,b).
The closure of (a,b] is similar except this time we have b to work with so we can use b to as close as possible to b, so the closure of (a,b] is [a,b].
