Closure of a left-bounded interval I am trying to figure out what I might be doing wrong in this problem: Let there be a collection $\mathcal{A}$ of intervals such that $\mathcal{A} = \{(a, \infty) \mid a \in \mathbb{R}\} \cup \{\varnothing\} \cup \{\mathbb{R}\}$. It is easy to show that this is a topology on $\mathbb{R}$. Find the closure of a set $A \subset \mathbb{R}$. 
Here is how proceed: Any element in this topology is of the form $(a, \infty)$. So, the closure of this set is $[a, \infty]$. Is this correct?
Regards,
RD

The closed sets are $\{\mathbb R-C\mid C \in \mathcal{A}\}=\{\mathbb R, \varnothing \}\cup\{(-\infty,a]\mid a\in R\}$
So for $A\subset \mathbb R$ the closure would be the inclusion of all closed sets containing $A$, namely $(-\infty,\sup A]$, when $\sup A$ can be equal to $\infty$, and in that case we'll consider $(-\infty,\infty]=\mathbb R$
