Defining open sets/neightbourhoods without balls I had a serious talk if it is possible to define the open sets of a topology without being in a metric space $(X,d)$ and a little bit more consistent that having a topology $(X,\tau)$. Then you have your open sets but the question is:
How to define open sets without the help of using balls? This could be with neightbourhoods. But this definition is requiring a metric space because a neightbourhood of a point is just a set $V$ for which there exist a ball contained in $V$. (you can argue that in general the open sets are defined via the basis of a topology which could be the balls). Can I define the neightbourhood in a more general way via topology, thus not being in a metric space.
In this scenario one could pick more basis for the topology which could be via bijection with homeomorphism of the open sets to another ones. For example deforming the open sets or the balls to more peculiar open sets.
 A: I don't know whether the following is what you're asking for, but here it is anyway. Start with any set $X$ and any collection $\mathcal S$ of subsets of $X$. Let $\mathcal B$ be the collection of all intersections of finite subfamilies of $\mathcal S$ (including the intersection $X$ of the empty subfamily of $\mathcal S$). Then define $\mathcal T$ to be the family of all unions of subfamilies of $\mathcal B$ (including the union $\varnothing$ of the empty subfamily of $\mathcal B$). Then $\mathcal T$ is a topology, i.e., we can take $\mathcal T$ to be the collection of open subsets of $X$ and thereby define a legitimate notion of "open".
In this situation, $\mathcal T$ is the smallest topology (i.e., the fewest open sets) that includes $\mathcal S$. One says that $\mathcal S$ is a subbasis for $\mathcal T$, that $\mathcal B$ is a basis for $T$, and that $\mathcal T$ is the topology generated by $\mathcal S$.
A: One way for $\mathbb R^n$ would be to first define the topology on $\mathbb R$ via the order (it's still the basis of open intervals, but now no longer interpreted as one-dimensional open balls), and then use the product topology on $\mathbb R^n$.
More exactly, you take the order of $\mathbb R$ (that is, the relation $a<b$) as given, and define the basis of the topology as all the open intervals, defined as $(a,b) = \{x\in\mathbb R:a<x<b\}$.
The topology on $\mathbb R^n$ then is the product topology, which for finite products ($\mathbb R^n$ has $n$ factors) is just the topology generated by the products of open sets of the factors.
