As we go on, we’re going to want to focus right in on a point in a topological space . We’re interested in the subsets of in which we could “wiggle around a bit” without leaving the subset. These subsets we’ll call “neighborhoods” of .
More formally, a neighborhood of is a subset of which contains some open subset , which contains the point . Then we can wiggle within the “nearby” set and not leave . Notice here that I’m not requiring itself to be open, though if it is we call it an “open neighborhood” of .
In fact, any open set is an open neighborhood of any of its points , since clearly it contains an open set containing — itself! Similarly, a subset is a neighborhood of any point in its interior. But what about a point not in its interior? If we take a point , but , then is a neighborhood of if and only if there is an open set with . But then since is an open set contained in , we must have , which would put into the interior as well. That is, a set is a neighborhood of exactly those points in its interior. In fact, some authors use this condition to define “interior” rather than the one more connected to orders.
So the only way for a set to be a neighborhood of all its points is for all of its points to be in its interior. That is, . But, dually to the situation for the closure operator, the fixed points of the interior operator are exactly the open sets. And so we conclude that a set is open if and only if it is a neighborhood of all its points — another route to topologies! We say what the neighborhoods of each point are, and then we define an open set as one which is a neighborhood of each of its points.
But now we have to step back a moment. I can’t just toss out any collection of sets and declare them to be the neighborhoods of . There are certain properties that the collection of neighborhoods of a given point must satisfy, and only when we satisfy them will we be able to define a topology in this way. Let’s call something which satisfies these conditions (which we’ll work out) a “neighborhood system” for and write it .
First of all (and almost trivially), each set in must contain . We’re not going to get much of anywhere if we don’t at least require that.
If is a neighborhood of , and , then must be a neighborhood as well since it contains whatever open set satisfies the neighborhood condition . Also, if and are two neighborhoods of then . That is, there must be a neighborhood contained in both and . We can sum up these two conditions by saying that is a “filter” in the partially-ordered set .
So, given a topology on we get a filter for each point. Conversely, if we have such a choice of a filter at each point, we can declare the open sets to be those so that implies that .
Trivially, satisfies this condition, as it doesn’t have any points to be a neighborhood of. The total space satisfies this condition because it’s above everything, so it’s in every filter, and thus is a neighborhood of every point.
Now let’s take two sets and , which are neighborhoods of each of their points, and let’s consider their intersection and a point . Since is in both and , each of them is a neighborhood of , and so since is a filter we see that must be a neighborhood of as well. We can extend this to cover all finite intersections.
On the other hand, let’s consider an arbitrary family of sets, each of which is a neighborhood of each of its points. Now, given any point in the union it must be in at least one of the sets, say . Now tells us first that by assumption, and then that by the filter property of . Thus we can take arbitrary intersections, and so we have a topology.
One caveat here: I might be missing something. Other definitions of a neighborhood system tend to include something along the lines of saying that every neighborhood of a point contains another neighborhood in its interior. I seem to have come up with a topology without using that assumption, but I’m willing to believe that there’s something I’ve missed here. If you see it, go ahead and let me know.