How to find closure of the set $A = \{(-3,-1), (-2,-2), (-1,-3)\}$, if topology looks like $\tau = \{(n,\infty)\times(n,\infty): n \in \mathbb{N} \} \,\cup\, \{ \emptyset,\mathbb{R}^{2} \}$?
I started to do it like this, here all closed sets:
1) $ \mathbb{R}^2 -(n,\infty)\times(n,\infty)$
2) $\mathbb{R}^2$
3) $\emptyset$
Closed sets containing $A$ are $\mathbb{R}^2 -(n,\infty)\times(n,\infty)$ and $\mathbb{R}^2$.
So, intersection of all closed sets looks like $\mathbb{R}^2 -(n,\infty)\times(n,\infty)$ and is it answer???
Definition: The closure of a set $X$ is the intersection of all closed sets $Y$ such that $X\subset Y.$
In this case we observe from the topology that: All closed sets are of the form $Y_n = \Bbb R^2 - (n,\infty)\times(n,\infty),$ and that $Y_n\subset Y_m$ if and only if $n\le m.$ Therefore we can write down a specialised definition of closure:
The closure of a set $X$ under this special topology is the set $Y_n$ such that $X \subset Y_n$ and $n$ is minimised.
So now let’s look at your points and decide what constraints they put on $n$:
$$ \begin{align} (-3,-1)&& n\ge -3\\ (-2,-2)&& n\ge -2\\ (-1,-3)&& n\ge -3 \end{align} $$
So we deduce that our closure must be $C=Y_{-2} =\Bbb R^2 - (-2,\infty)\times(-2,\infty),$ and to check this we could show that: 1) $C$ is closed, 2) all the points from our set are contained in $C$, 3) any other closed set is either a subset of $C$ or does not contain all the points.