Why do we only allow for finite intersections in the definition of a topology?

Question

I have read the strongly related questions ("Why do we require a topological space to be closed under finite intersection?", "For the definition of a topological space, why is the internal union allowed to be infinite, while the internal intersection is restrained to be finite?", "Topology definition: finite intersections vs. infinite unions" and "Why for unions we can have arbitrary number of sets and for intersections it's only finite?") and most of those threads answer this question either vaguely by stating that the resulting topology wouldn't be very interesting (as the standard topology is reduced to the discrete topology of R) or via neccessity by showing that the topology would then contain closed sets as well (Although via the topological definition of an open set these would still be open).

However, I would like to build upon the first of the answers stated above. In M. Nakahara's book "Geometry, Topology and Physics", there is an exercise on page 49 regarding this very issue:

"Exercise 2.25 In definition 2.23, axioms (ii) [closure with respect to unions] and (iii) [closure with respect to intersections] look somewhat unbalanced. Show that, if we allow infinite intersection in (iii), the usual topology in $\mathbb{R}$ reduces to the discrete topology (thus not very interesting)."

While I was able to show this for $\mathbb{R}$, I was not able to show this for any arbitrary topology/set. My question now is whether any topology is reduced to the discrete topology, if inifinite intersections are allowed? If so, then this would explain why these topologies aren't "interesting".

Answer 1

As Martin Brandenburg comments, a topology in which the intersection of every family of open sets is open is known under the name Alexandrov topology.

By definition each Alexandrov topology is a topology, but the converse is not true. Note, however, that the discrete topology is always an Alexandrov topology. But as you say, this topology is not very interesting.

To each topology $\tau$ one can assign the Alexandrov topology $\tau^*$ generated by $\tau$. This is the coarsest Alexandrov topology containing $\tau$ (i.e. the intersection of all Alexandrov topologies containing $\tau$). Thus $\tau^*$ is always finer than $\tau$.

When is $\tau^*$ the discrete topology?

Fact 1. An Alexandrov topology $\vartheta$ is discrete if and only if the space $(X,\vartheta)$ is a $T_1$-space.

Trivially each discrete space is $T_1$. Conversely, if $(X,\vartheta)$ is $T_1$ (which is equivalent to all one-point subsets being closed), then all subsets are closed because they are unions of one-point subsets. Note that all unions of closed subsets are closed in Alexandrov topologies. This means that all subsets are open, i.e. $\vartheta$ is discrete.

Fact 1 shows that Alexandrov topologies are not very interesting if we require a very mild separation axiom.

Fact 2. If a topological space $(X,\tau)$ is $T_1$ , then $(X,\tau^*)$ is $T_1$ and hence discrete.

This is true because $\tau^*$ is finer than $\tau$.

The two above facts show that it would not be a good idea to adopt closedness under arbitrary intersections of open sets as a general axiom for topological spaces. For example, none of the standard spaces occuring in analysis would be a topological space in that sense.

Answer 2

Since finite, non-discrete topologies exist, what you want cannot happen. For an extreme case, take any set $X$ with two or more elements, and consider the topology $\{\varnothing, X\}$.

That said, most (not all, by any means, but most) topologies people care about are Hausdorff. The reason being that it is very reasonable to expect your topology to separate points (though, there are weaker ways to separate points than Hausdorff). In a Hausdorff topology the argument you used for $\mathbb R$ will work, and you can show that any singleton is the intersection of all open sets that contain it.