The Unapologetic Mathematician

Mathematics for the interested outsider


As we go on, we’re going to want to focus right in on a point x in a topological space (X,\tau). We’re interested in the subsets of X in which we could “wiggle x around a bit” without leaving the subset. These subsets we’ll call “neighborhoods” of x.

More formally, a neighborhood N of x is a subset of X which contains some open subset U\subseteq X, which contains the point x. Then we can wiggle x within the “nearby” set U and not leave N. Notice here that I’m not requiring N itself to be open, though if it is we call it an “open neighborhood” of x.

In fact, any open set U is an open neighborhood of any of its points x, since clearly it contains an open set containing x — 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 x\in S, but x\notin S^\circ, then S is a neighborhood of X if and only if there is an open set U with x\in U\subseteq S. But then since U is an open set contained in S, we must have U\subseteq S^\circ, which would put x 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, S^\circ=S. 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 S 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 x. 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 x and write it \mathcal{N}(x).

First of all (and almost trivially), each set in \mathcal{N}(x) must contain x. We’re not going to get much of anywhere if we don’t at least require that.

If S\in\mathcal{N}(x) is a neighborhood of x, and S\subseteq T, then T must be a neighborhood as well since it contains whatever open set U satisfies the neighborhood condition x\in U\subseteq S. Also, if S and T are two neighborhoods of x then x\in S^\circ\cap T^\circ=(S\cap T)^\circ\subseteq S\cap T. That is, there must be a neighborhood contained in both S and T. We can sum up these two conditions by saying that \mathcal{N}(x) is a “filter” in the partially-ordered set P(X).

So, given a topology \tau on X we get a filter \mathcal{N}(x) 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 U so that x\in U implies that U\in\mathcal{N}(x).

Trivially, \varnothing satisfies this condition, as it doesn’t have any points to be a neighborhood of. The total space X 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 U and V, which are neighborhoods of each of their points, and let’s consider their intersection and a point x\in U\cap V. Since x is in both U and V, each of them is a neighborhood of x, and so since \mathcal{N}(x) is a filter we see that U\cap V must be a neighborhood of x as well. We can extend this to cover all finite intersections.

On the other hand, let’s consider an arbitrary family U_\alpha of sets, each of which is a neighborhood of each of its points. Now, given any point x in the union \bigcup\limits_\alpha U_\alpha it must be in at least one of the sets, say U_A. Now x\in U_A\subseteq\bigcup\limits_\alpha U_\alpha tells us first that U_A\in\mathcal{N}(x) by assumption, and then that \bigcup\limits_\alpha U_\alpha\in\mathcal{N}(x) by the filter property of \mathcal{N}(x). 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 x 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.

November 15, 2007 - Posted by | Point-Set Topology, Topology


  1. Yes, you do get a topology without that assumption, but you’d then lose the one to one correspondence between topologies as usually defined and topologies via neighborhood filters. For example, take X = \{0, 1, 2\}, and define a filter at each point by: N_0 = \{\{0, 1\}, X\}, N_1 = \{\{1, 2\}, X\}, N_2 = \{\{0, 2\}, X\}. Then the only open sets are the empty set and $X$; in particular, you can’t retrieve the N_i from the topology in this example.

    Comment by Todd Trimble | November 15, 2007 | Reply

  2. Okay, so many different neighborhood systems will give the exact same topology, but only the one that satisfies the compatibility condition between the different neighborhood filters will be the system we can get from a topology, right?

    Can we get the “right” filter by just throwing out all the “bad” neighborhoods, which don’t contain a neighborhood in their interiors?

    Comment by John Armstrong | November 16, 2007 | Reply

  3. Yes to the first question, and I don’t see anything wrong with the formulation of the second (i.e., I don’t think there are fatal objections on grounds of impredicativity or anything like that). Anyway, I think it’s clear to both of us category wonks what’s going on: there’s an “underlying” functor which maps neighborhood systems (in the sense of your post) to topologies, and this has a left adjoint which maps a topology to the neighborhood system it generates, and throwing out the bad neighborhoods amounts to stabilizing, i.e., taking the coclosure (wrt the induced comonad) of the neighborhood system.

    All this reminds me of the true story of an undergraduate student of the high-powered Australian categorist Ross Street who, when asked in an oral exam if he could define the notion of topological space, answered accurately but bewilderingly,

    “Why yes, it’s just a relational \beta-module!”

    There is some such ultra-fancy definition, which I believe was worked out in the mid-sixties by Michael Barr; I’ve never worked out what it’s saying but I think it must be somewhere in the neighborhood (ha ha) of what we’re talking about now. (This \beta presumably refers to the monad on Set whose algebras are compact Hausdorff spaces, viz., \beta(X) is the set of ultrafilters on X). Might be fun to work out some time.

    Comment by Todd Trimble | November 16, 2007 | Reply

  4. Oh, it was Ross Street’s student who had that topological space definition! I’ve had the ghost of that anecdote running in my head for almost half a year now.

    Comment by Mikael Johansson | November 16, 2007 | Reply

  5. That you don’t need the usual assumption that neighbourhoods contain other neighbourhoods in their `interiors’, but that the `bad’ neighbourhood systems are not in the image of the adjoint functor is very similar to what happens for non-idempotent closure operators. And its seems that going back and forth between `bad’ neighbourhood systems and `bad’ closure operators, one doesn’t throw out anything.

    Comment by Benoit Jubin | November 18, 2007 | Reply

  6. Benoît, that’s a great observation.

    I may have mentioned it before, but I haven’t really looked at the basics of topology in so long precisely because it was always just glossed over swiftly as another commenter suggested it should be. Now that these posts bring me back around to look at it directly, there’s a lot of really fascinating stuff in here. It’s doubly bizarre to find it all over again because I’m a topologist myself!

    Comment by John Armstrong | November 18, 2007 | Reply

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