Why the radius of the ball needs to be rational? Second countability

Question

In proving that $\mathbb{R}$ is second countable with the usual metric, one says that each member of the family of open balls $\mathcal{B}$ is centered in each rational $x\in V$ for $V\subseteq \mathbb{R}$ with rational radius. Examples are given here and here. To fix the center as a rational is needed to see that $V = \cup_x B_r(x)$ is countable.

Why the radius needs to be rational? For example in $(0,\pi)$, there can be balls with radius proportional to a fraction of $\pi$, hence irrational radius.

A topological space $X$ is said to be second countable if there exists some countable family $\mathbb{B}$ of open sets such that each open set $V\subseteq X$ can be expressed as the union of some subfamily of $\mathbb{B}$

Answer 1

EDIT: per the OP's comment to ndhanson3's answer, it seems the sticking point is:

Why does the set $A$ of balls with rational center and arbitrary (positive) radius not demonstrate the second countability of $\mathbb{R}$ (with the usual topology)?

Well, we have to check two things. First, we have to show that $A$ is in fact a base for the topology in question. This is a bit tedious to do, but it isn't hard, so let's just assume we've done it.

The issue however is the countability requirement: to show that $\mathbb{R}$ is second-countable we need a countable base, not just an arbitrary base. And when we say "countable base" we really do mean "countable base:" the base $A$ above may "feel countable" (in some sense there are countably many "types of" element of $A$), but since it's not literally a countable set it doesn't help us.

(Why isn't $A$ countable? Well, think about the set $\{B_r(42): r\in\mathbb{R}_{>0}\}$. This is a subset of $A$, and it's "as big as" the set of positive reals; now, is the set of positive reals countable or uncountable?)

By contrast, since "countable times countable equals countable" if we look at a set of balls with rational centers and rational radii, we do indeed wind up with a countable set - since the set of positive rational numbers (unlike the set of positive real numbers) is countable.

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OK, why specifically the rationals?

This is indeed unnecessary - it's just a convenient choice.

All we actually need to show second-countability is:

• A countable dense set $P$ of points in our space.

• A countable set $D$ of positive real numbers which is "downwards-dense" in the sense that for each $\epsilon>0$ there is some $\delta\in D$ with $\epsilon>\delta$.

As soon as we have such $P$ and $D$, we can show that our space is second-countable: the set of balls with radii in $D$ and with center in $P$ is a countable base of our space. (Moreover, note that $D$ doesn't depend on our space at all: we can fix a $D$ to use in every situation, so that a metric space is second-countable if it has a countable dense set. This is not true for general topological spaces.)

At this point it's just a question of what's most natural for you. For spaces "based on" $\mathbb{R}$, the usual choice of $P$ is something "based on" $\mathbb{Q}$ (e.g. $\mathbb{Q}$ itself for $\mathbb{R}$, $\mathbb{Q}^2$ for $\mathbb{R}^2$, etc.). Meanwhile, the usual choices for $D$ that I've seen are $\mathbb{Q}_{>0}$, $\{{1\over n}: n\in\mathbb{N}\}$, and $\{2^{-n}: n\in\mathbb{N}\}$ (the former is by far the most common in my experience but every so often one of the latter two is a bit more convenient). But we could also use something like $\{q\pi: q\in\mathbb{Q}_{>0}\}$: it just has to be countable and downwards-dense.

Answer 2

Sure, you could use radii that are rational multiples of $\pi$ as well. It doesn't matter. You want to prove that a countable basis exists, so the most natural countable basis to find is the one with rational radii.