# The Unapologetic Mathematician

## Neighborhoods

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