Concerning the closure of a set in this topology

It is not hard to see that the set $\mathcal{I} = \{ (a, \infty) \vert a \in \mathbb{R} \} \cup \{\mathbb{R}\} \cup \{\emptyset\}$ is a basis for a topology of semi-infinite intervals.

Given some arbitrary element of this basis (that is neither the empty set nor the whole space), I would like to find its closure. Our set is trivially open, so it is equal to its interior, and thus we simply need the boundary points to see its closure.

The nature of our basis elements makes it clear that the point $a$ would be the one point on the boundary of said element (since all open sets that contain it have nontrivial intersection with both the interior of our element and $ \mathbb{R}$ sans this element). Naturally, I would believe that the closure of my basis element is $ [a, \infty)$. However, on complementation we obtain the set $ (- \infty, a)$, which is not open in our topology since, I claim, it cannot be obtained through an arbitrary union of basis elements.

This would lead me to believe that the true closure of any given basis element is simply the whole space since its complement is trivially open and also contains the boundary point in question. Is this the case? How could I have seen this sooner, if so? Also if so, why does it seem to be the case here that the closure is not the interior with the boundary? Furthermore, would the closure of any open set in this topology also be the whole space, assuming it is the case for our basis elements as I have explained?

These questions are quite basic, but I have no mathematicians nearby to dispel my confusion. Any and all help appreciated.

Edit: Thank you for your help. I understand the problem and concept much better now.

Answer

Notice that the family you've written is not just a basis for a topology, but the topology itself.
With this in mind, one should start by looking at the class of closed subsets, which is $$\{(-\infty,a]\,:\,a\in\Bbb R\}\cup\{\Bbb R,\emptyset\}$$

and then notice that the only one containing a non-empty open set is $\Bbb R$. So, every non-empty open set is dense. Finally, the boundary of $(x,\infty)$ is in fact $(-\infty,x]$.