What is more general than a topological space? I am not sure if this question actually makes sense, and I am also not sure if the details of this question are entirely correct, so any corrections would be greatly appreciated!
In my math class, we learned that every inner product space is a normed space and that every normed space is a metric space. Is it true that every metric space is a topological space? But, the converses of these may not be true. Maybe I should replace "is" with "may induce a"?
But since these are getting more and more general, is there something that a topological space may induce or something that is more general than a topological space?
 A: I will try to answer your question in elementary terms. You have seen that every 
normed space is a metric space (but the converse is no true) and that every metric space is a topological space (but the converse is no true). Now you are asking if there is some kind of space, let's call it a ??-space such that every topological space is a ??-space (but the converse is no true).
Of course, you could just take a ??-space to be a set, but there is a slightly more subtle answer. You could take as ??-space a preordered set. Let me first recall that a preorder is a reflexive and transitive relation. It turns out that every topological space is naturally equipped with a preorder, called the specialization preorder, which can be defined as follows:
$$
x \leqslant y \iff \overline{\{x\}} \subseteq \overline{\{y\}},
$$
where $\overline{S}$ denotes the closure of a set $S$. However, you might be disappointed by this answer, because for an Hausdorff space (also called $T_2$-space), the specialization preorder is the equality relation. On the positive side, it the space is Kolmogorov (also called $T_0$), then the specialization preorder is a (partial) order.
In the language of categories, I just described the forgetful functor from the category of topological spaces to the category of preordered sets.
A: Pretopology spaces, nhood spaces, precloser spaces.
A precloser pcl, operator lacks plc plc A = pcl A.
