Urysohn Metrization Theroem : Proof of Injectivity I am reading Munkres Topology and following given Urysohn metrization theorem and its proof, but can't understand why the injectivity simply given after define a index functions and product them.
I had highlighted the point that I am missing at the below inserted images. 

 A: $F(x)=F(y)$ if and only if $f_n(x) = f_n(y)$ for all $n.$ 
The key fact they are using is that there is an $n$ such that $f_n(x) > 0$ and $f_n(y)=0.$ (This follows from the first step of the proof and the fact that regularity implies there is a neighborhood of $x$ that doesn't contain $y$.) This means there is an $n$ such that $f_n(x)\ne f_n(y),$ so $F(x)\ne F(y)$
A: Flip back to page 195 and recall that the definition of a regular set $X$ assumes that one-point sets are closed in $X$. Further, $X$ is regular if for each pair consisting of a point x and a closed set $B$ disjoint from $x$, there exist disjoint open sets containing $x$ and $B$, respectively.
How can we use that to show $F$ is injective? $F$ is injective if $x \neq y$ implies $F (x) \neq F(y)$. So suppose $x, y \in X$ with $x \neq y$. Let $B = \{y\}$, then $B$ is closed in $X$. Since $X$ is regular, there exist disjoint neighborhoods $U$ and $V$ of $x$ and $y$ respectively.
Now we can apply step 1 of the proof. It implies that there exists an $n$ such that $f_n (x) > 0$ and $f_n = 0$ outside of $U$. But note $y \not \in U$, so $f_n (y) = 0$. So some component of $F$, namely $f_n$, has different values at $x$ and $y$, so $F(x) \neq F(y)$. Of course, since $x$ and $y$ were arbitrary, this shows that $F$ is injective.
