# Defining a topology on a (non-metric) space…but what is the definition of 'open'?

A topology on some set is defined as a collection of open subsets satisfying three axioms. I was first introduced to the concept of 'open' in the context of a metric space where, for any point in an open subset, there exists an open ball contained in that subset.

Does this new definition of 'open subsets' in the topology shatter my previous intuition? It's a very general definition and seems arbitrary rather than intuitive. Take, for example, $\mathbb{R}^2$. Say we decide to include a single point, the origin, in the topology. So this is now an 'open subset' despite being an isolated point of the space?

Thank you.

• If you check the definition, you’ll find that for any set $X$ the power set $\wp(X)$ is a topology on $X$; it’s called the discrete topology, and yes, in that topology each singleton set $\{x\}$ with $x\in X$ is an open set whose only element is an isolated point of the space. – Brian M. Scott Sep 6 '16 at 19:18
• Be careful when you say "new definition of open subset." The definition of "topology" does not define what an open set is, but instead requires that a topological space define what "open" means. Then, to be a topology, that definition of "open" must satisfy the three axioms. It is like how in the definition of "vector space," "vector" is left free -- there is no sense in asking what a general vector is, since it's up to circumstance. – Kyle Miller Sep 6 '16 at 19:25
• I removed the "differential geometry" and "differential topology" tags. This question is unrelated. – Fly by Night Sep 6 '16 at 19:39

Before you give the set $\mathbb R^2$ a topology, it is just a set of points. You shouldn't think of it as having any geometric structure. The fact that you think of $\mathbb R^2$ as a plane is that "ordinarily", the topology on $\mathbb R^2$ is the one which is given by any of the common metrics on $\mathbb R^2$.

You could make a new topology on $\mathbb R^2$ in which all old open sets are still open, but $\{0\}$ is also open. The way you should think of this space geometrically is as follows: it is a plane with a single point missing (the set $\mathbb R^2\setminus \{0\}$) together with a separate point floating beside it (the set $\{0\}$).

Edit: in fact, this topology is also given by a metric: if $d$ is any ordinary metric on $\mathbb R^2$, then the metric $d': \mathbb R^2 \times \mathbb R^2 \to \mathbb R$ which is given for $x,y \neq 0$ by $d'(x,y) = d(x,y)$ and by $d'(x,0) = d(x,0) + 1$ induces this topology.

Does this new definition of 'open subsets' in the topology shatter my previous intuition?

Yes, in a way. Intuition is often the motivation for generalization, then needs to be refined. The topology on a metric space comes from the intuition you have for how distances behave in Euclidean space. There are useful topologies that don't come from metrics - there your intuition is guided by how open sets behave in metric spaces.

One such example is the Zariski topology. You'll find others here: https://mathoverflow.net/questions/52032/examples-of-non-metrizable-spaces .

"The new definition" should not shatter your intuition completely, and one should always bear in mind how topology of metric spaces works to have some intuition about general topological spaces. One should also be careful what properties of metric spaces fail in general topological space.

Open sets are precisely those that satisfy the three axioms. Now, in metric spaces, open set is union of open balls. It is good exercise to prove:

$U$ is union of open balls if and only if $(\forall x\in U)(\exists r>0)\ B(x,r)\subseteq U$.

Family of open balls in a metric space forms what is known as base of topology. It is very convenient to define topology in terms of it's base - you take some subsets which you want to be open, and generate the whole topology with it (this family cannot be chosen completely freely, as mentioned in the link above). This is precisely how we define topology on metric space: we take open balls to be it's base.