I am reading Proposition 2.1.4 in Dudley's "Real Analysis and Probability (Cambridge Edition)" but I have some troubles.
That proposition and the proof are here:
A metric space $(S,d)$ is second-countable iff it is separable.
Proof: ($\longleftarrow$) Let $A$ be countable and dense in $S$. That is $\overline A=S $. Let $\mathcal U$ be the set of all balls $B(x, 1/n)$ for $x\in A$ and $n=1,2,\dots$. To show that $\mathcal U$ is a base, let $U$ be any open set and $y\in U$. Then for some $m$, $B(y, 1/m)\subset U$. Take $x\in A$ with $d(x,y)<1/(2m)$ Then $y\in B(x, 1/(2m))\subset B(y, 1/m)\subset U$, so U is the union of the elements of $\mathcal U$ that it includes, and $\mathcal U$ is a base, which is countable. Since Cartesian products of countable sets (in this proof, $A$ and $\mathbb{N}$ are so) is also countable, $\mathcal U$ is a countable base.
($\longrightarrow$) Suppose there is a countable base $\mathcal V$ for the topology, which we may assume consists of nonempty sets. By the axiom of choice, let $f$ be a function on $\mathbb{N}$ whose range contains at least one point of each set in $\mathcal V$. Then this range is dense.
I have two questions:
-
in the proof of ($\longleftarrow$), where did Dudley use the denseness of A?
Note that in Theorem 2.1.3 in his book says that for any $A\subset S$, $\overline A$ is the set of all $x\in S$ such that for some net $x_i \longrightarrow x$ with $x_i\in A$ for all $i$.
That is, $\overline A$ is the set of all $x\in S$ such that any neighborhood of $x$ contains all but finitely many points $x_i$ with $x_i\in A$. Do I need to use this fact? -
in the proof of ($\longrightarrow$), why the range of $f$ is dense?
Best Answer
Both have the same reason: $A$ is dense in $X$ iff for all open balls $B(x,r), x\in X, r>0$ we have $A \cap B(x,r) \neq \emptyset$.
He uses this in $\longrightarrow$: if $B(x,r)$ is some ball, it contains some $V \in \mathcal{V}$ and then $f(V) \in V $ intersects that ball.
In $\longleftarrow$ he says: "pick $x \in A$ with $d(x,y) < \frac{1}{2m}$", which is picking a member of $A \cap B(y, \frac{1}{2m})$, so it applies the denseness of $A$ (intersecting all balls).