# The Unapologetic Mathematician

## Countability Axioms

Now I want to toss out a few assumptions that, if they happen to hold for a topological space, will often simplify our work. There are a lot of these, and the ones that I’ll mention I’ll dole out in small, related collections. Often we will impose one of these assumptions and then just work in the subcategory of $\mathbf{Top}$ of spaces satisfying them, so I’ll also say a few things about how these subcategories behave. Often this restriction to “nice” spaces will end up breaking some “nice” properties about $\mathbf{Top}$, and Grothendieck tells us that it’s often better to have a nice category with some bad objects than to have a bad category with only nice objects. Still, the restrictions can come in handy.

First I have to toss out the concept of a neighborhood base, which is for a neighborhood filter like a base for a topology. That is, a collection $\mathcal{B}(x)\subseteq\mathcal{N}(x)$ of neighborhoods of a point $x$ is a base for the neighborhood filter $\mathcal{N}(x)$ if for every neighborhood $N\in\mathcal{N}(x)$ there is some neighborhood $B\in\mathcal{B}(x)$ with $B\subseteq N$. Just like we saw for a base of a topology, we only need to check the definition of continuity at a point $x$ on a neighborhood base at $x$.

Now we’ll say that a topological space is “first-countable” if each neighborhood filter has a countable base. That is, the sets in $\mathcal{B}(x)$ can be put into one-to-one correspondence with some subset of the natural numbers $\mathbb{N}$. We can take this collection of sets in the order given by the natural numbers: $B_i$. Then we can define $U_0=B_0$, $U_1=U_0\cap B_1$, and in general $U_n=U_{n-1}\cap B_n$. This collection $U_i$ will also be a countable base for the neighborhood filter, and it satisfies the extra property that $m\geq n$ implies that $U_m\subseteq U_n$. From this point we will assume that our countable base is ordered like this.

Why does it simplify our lives to only have a countable neighborhood base at each point? One great fact is that a function $f:X\rightarrow Y$ from a first-countable space $X$ will be continuous at $x$ if each neighborhood $V\in\mathcal{N}_Y(f(x))$ contains the image of some neighborhood $U\in\mathcal{N}_X(x)$. But $U$ must contain a set from our countable base, so we can just ask if there is an $i\in\mathbb{N}$ with $f(B_i)\in V$.

We also picked the $B_i$ to nest inside of each other. Why? Well we know that if $f$ isn’t continuous at $x$ then we can construct a net $x_\alpha\in X$ that converges to $x$ but whose image doesn’t converge to $f(x)$. But if we examine our proof of this fact, we can look only at the base $B_i$ and construct a sequence that converges to $x$ and whose image fails to converge to $f(x)$. That is, a function from a first-countable space is continuous if and only if $\lim f(x_i)=f(\lim x_i)$ for all sequences $x_i\in X$, and sequences are a lot more intuitive than general nets. When this happens we say that a space is “sequential”, and so we have shown that every first-countable space is sequential.

Every subspace of a first-countable space is first-countable, as is every countable product. Thus the subcategory of $\mathbf{Top}$ consisting of first-countable spaces has all countable limits, or is “countably complete”. Disjoint unions of first-countable spaces are also first-countable, so we still have coproducts, but quotients of first-countable spaces may only be sequential. On the other hand, there are sequential spaces which are not first-countable whose subspaces are not even sequential, so we can’t just pass to the subcategory of sequential spaces to recover colimits.

A stronger condition than first-countability is second-countability. This says that not only does every neighborhood filter have a countable base, but that there is a countable base for the topology as a whole. Clearly given any point $x$ we can take the sets in our base which contain $x$ and thus get a countable neighborhood base at that point, so any second-countable space is also first-countable, and thus sequential.

Another nice thing about second-countable spaces is that they are “separable”. That is, in a second-countable space $X$ there will be a countable subset $S\subseteq X$ whose closure $\mathrm{Cl}(S)$ is all of $X$. That is, given any point $x\in X$ there is a sequence $x_i\in S$ — we don’t need nets because $X$ is sequential — so that $x_i$ converges to $x$. That is, in some sense we can “approximate” points of $X$ by sequences of points in $S$, and $S$ itself has only countably many points.

The subcategory of all second-countable spaces is again countably complete, since subspaces and countable products of second-countable spaces are again second-countable. Again, we have coproducts, but not coequalizers since a quotient of a second-countable space may not be second-countable. However, if the map $X\rightarrow X/\sim$ sends open sets in $X$ to open sets in the quotient, then the quotient space is second-countable, so that’s not quite as bad as first-countability.

Second-countability (and sometimes first-countability) is a property that makes a number of constructions work out a lot more easily, and which doesn’t really break too much. It’s a very common assumption since pretty much every space an algebraic topologist or a differential geometer will think of is second-countable. However, as is usually the case with such things, “most” spaces are not second-countable. Still, it’s a common enough assumption that we will usually take it as read, making explicit those times when we don’t assume that a space is second-countable.

November 28, 2007 - Posted by | Point-Set Topology, Topology

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2. [...] This is a metric space, with a metric function defined by . This is nice because metric spaces are first-countable, and thus sequential. That is, we can define the topology of a (semi-)normed vector space by [...]

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4. [...] compact closure — the same is true of . We don’t however, know that is Hausdorff or second-countable, both of which we’ll [...]

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