Why do we want topologies to be closed under both finite and infinite union but not infinite intersection?

Question

So I recently read the definition of a topological space and a topology from a book, and according to it, the topology must be closed under finite and infinite union, but it must only be closed under finite intersection. Clearly there is difference between requiring topologies to be closed under finite union and requiring them to be closed under both finite and infinite union, this is because imagine the topology is made up of infinite sets and every set sort of "adds" a new element to the infinite union, then there won't be any finite union that is equal to the infinite one, and thus some topologies might be closed under finite union only, but not under infinite union.

At first I thought maybe the reason why they definition doesn't explicitly say that it should be closed under infinite intersection was because it wouldn't change anything, as in, every topology that is closed under finite intersection will also be closed under infinite intersection, however I think I found a counterexample. Consider the set $X = \mathbb{N}$ and the topology $\tau = \{\emptyset, \mathbb{N}, (\mathbb{N} \setminus \mathbb{N}_2) \cup \{1\}, (\mathbb{N} \setminus \mathbb{N}_3) \cup \{1\}, (\mathbb{N} \setminus \mathbb{N}_4) \cup \{1\}, ... \}$ where we define $\mathbb{N}_i$ as the set $\{1, 2, 3,..., i\}$. This is a topology by the definition the book uses since it contains $X$ and $\emptyset$ and it's also closed under finite and infinite union as well as closed under finite intersection, however it is not closed under infinite intersection since it doesn't contain the set $\{1\}$.

So I have two questions, is my reasoning so far correct? and also, if it is, then why do we want the topology I showed before to be a topology and why not make a stronger requirement that it also needs to be closed under infinite intersection?

Answer 1

There are topologies that are closed under arbitrary intersections. These are known as Alexandrov spaces: https://topology.pi-base.org/properties/P000090

Unfortunately, there's another very common topological property: "a single point set is closed", a.k.a. $T_1$. Examples of such spaces are metrizable spaces, spaces whose open sets are defined by sets of points that are within a certain distance of other points.

So what's wrong with that? Well, if you put these two properties together, then you can prove that any such space is discrete, that is, every set is both open and closed: https://topology.pi-base.org/theorems/T000267

At that point, what's the point of topology if everything is open and everything is closed? So we usually don't want to assume that any intersection of open is open.

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One other perspective: the intuition of an open set is that it should be a "neighborhood" of the sets it contains, that it, it has at least a bit of space or wiggle room around each of its points. For example, $(-1/N,1/N)$ is an open neighborhood of $0$ in the usual topology of the real numbers, no matter how big $N$ is.

But if we can intersect as many open sets as we want, then we must have $\bigcap\{(-1/N,1/N):N\in\mathbb Z^+\}=\{0\}$ open as well. But that doesn't provide the desired "wiggle room" around 0 we expect in the usual topology.

Answer 2

why not make a stronger requirement that it also needs to be closed under infinite intersection?

Mathematical definitions are not made to be strong. They're made to try to capture or abstract some type of behavior or object that seems like it's worth studying.

Very informally, an open set is one such that, no matter where you are within that set, you can still move around at least a little bit in any direction, and still remain in that set. If you allow infinite intersections of open sets to be open, then as the $\bigcap_{n = 1}^\infty (-1/n, 1/n)$ example shows, your definition doesn't abstract the desired type of behavior.

Answer 3

This is a good question, I am only a student of topology, but here is my answer. Much of general topology is concerned with the ideas of "open sets." This is a fundamental concept and it is used to define continuity, convergence, and so many more topological properties. Intuitively, the fact that a topology is closed under infinite unions allows one to construct all open sets from other open sets by "pasting" open sets together. However, closure under arbitrary intersections would allow you to create non-open sets from open sets. For instance, consider the Euclidean topology on $\mathbb{R}$ and take $\bigcap_{n \in \mathbb{N}} \left(-\frac{1}{n}, \frac{1}{n}\right) = \{0\}$. This is an infinite intersection and results in a non-open set.

Answer 4

Infinite intersections of neighbourhoods of a point tell me something about.. "infinitesimally close". Or maybe they allow that possibility. But if a neighbourhood is to allow some definite amount of free space about the point, this flies in the face of "infinitesimally close"; the two informal concepts contradict one another. And literally they cause a problem when you image metric spaces and the classic example already mentioned, of $(-1/n,1/n)$.

So, morally, this is not a suitable condition. And practically it's unsuitable because it ruins an enormous amount of important examples. However we demand arbitrary unions to be union because, well, if $A$ is my neighbourhood, $A\cup\{\text{anything else}\}$ shall also be my neighbourhood. I have introduced more space, but not taken any away; of course $A\cup\bigcup_iA_i$ should also be my neighbourhood, and for the same reasons it should be the neighbourhood of every point it contains, so the arbitrary union of open sets must (morally) be open.