We shall prove that X is metrizable by imbedding X into a metrizable space Y;that is by showing that X is homeomorphic to some subspace of Y.
Step 1: We prove the following:-
There exists a countable collection of continuous functions $f_n:X\rightarrow [0,1]$ having the property that given any point $x_0$ of $X$ and any neighborhood $U$ of $x_0$,there exists an index $n$ such that $f_n(x_0)>0$ and $f_n$ vanishes outside $U$.
Proof:
Let {$B_n$} be the countable basis for $X$.For each pair $n,m$ of indices for which $\overline{B_n} \subset B_m,$apply Urysohn's lemma to choose a continuous function $g_{n,m}:X\rightarrow[0,1] $ such that $g_{n,m}[\overline{B_n}]=\{1\}$ and $g_{n,m}[X-B_m]=\{1\}$. Then the collection {$g_{n,m}$} satisfies our requirement:
Given $x_0$ and given a neighbourhood $U$ of $x_0,$ one can choose a basis element $B_m$ containing $x_0$ that is contained in $U$.
Using Regularity,one can then choose $B_n$ so that $x_0 \in B_n$ and $\overline{B_n} \subset B_m$.Then $n,m$ is a pair of indices for which function $g_{n,m}$ is defined; and it is positive at $x_0$ and vanishes outside of $U$. Because the collection {$g_{n,m}$} is indexed with a subset of $\mathbb{Z}_+ \times \mathbb{Z}_+,$, it is countable; therefore it can be reindexed with the positive integers, giving us the desired collection $\{f_n\}$.
Proof of Urysohn's metrization theorem-
Given a function $f_n$ of step 1,take $\mathbb{R}^{\omega} $ in the product topology and define a map $F: X\rightarrow \mathbb{R}^{\omega}$ by the rule-
$F(x)=(f_1(x),f_2(x),f_3(x),f_4(x),\ldots)$
We assert that $F$ is an embedding-
$F$ is continuous
Because $\mathbb{R}^{\omega}$ has a product topology and each $f_n$ is continuous.
$F$ is injective
???????????
$F$ is a homeomorphism of $X$ onto its image, the subspace $Z=F[X]$ of $\mathbb{R}^{\omega} $.
Because each component of $F$ is continuous,so $F$ is continuous.(1.Is it correct reason?)
$F:X\rightarrow F[X]$ is a bijection.
So,we need only show that for each open set $U$ in $X,$ the set $F[U]$ is open in $Z=F[X]$.
Let $z_0\in F(U).$
We shall find an open set $W$ of Z such that $z_0\in W\subset F(U)$.
Let $x_0\in U$ such that $F(x_0)=z_0.$Choose an index $N$ for which $f_N(x_0)>0$
anf $f_N(X-U)=${$0$.}
Take the open ray $(0,\infty) $ in $\mathbb R$ and let $V$ be the open set $V=\pi_N^{-1}((0,\infty))$ of $\mathbb R^{\omega}$.
Let $W=V\cap Z;$ then $W$ is open in $Z$ by definition of subspace topology.We assert that $z_0\in W\subset F(U).$
First $z_0\in W$ because $\pi_N(z_0)=\pi_NF(z_0)=f_N(x_0)>0.$(2.PLEASE EXPALIN THIS!)
Second,$W\subset F(U).$For if $z\in W$ then $z=F(x)$ for some $x\in X$ and $\pi_N(z)\in (0,\infty)$.
Since,$\pi_N(z)=\pi_N(F(x))=f_N(x),$ and $f_N$ vanish outside $U,$the point $x$ must be in $U.$
Then,$z=F(x)$ is in $F(U)$,as desired.
Thus,$F$ is an imbedding of $X$ in $\mathbb R^{\omega}.$
Please,help me showing $F$ is injective,and justifications for (1)
and (2).
Also,examine this proof critically,and if there is some scope of improvement please let me know…
Best Answer
$F$ injective is quite obvious as $x \neq y$ implies that there is some basic element $B_n$that contains $x$ and misses $y$ (we include $T_1$ in regularity for this theorem). By regularity we have some $x \in \overline{B_m} \subseteq B_n$ and then when $k$ is the index corresponding to this $(m,n)$ pair, $f_k(x) = 1$ and $f_k(y) = 0$, ensuring that $F(x) \neq F(y)$, because the images differ at least in one coordinate.
$F$ is continuous as $\pi_k \circ F = f_k$ is continuous for all $k$, by the universal property the product topology.
In the $F$ being open proof, the step where we chose the $f_N$ is where the "magic" happens later; we have $x_0 \in U$ with $F(x_0) = z_0$. Then we find basic elements $(n,m)$ such that $x_0 \in \overline{B_n} \subseteq B_m \subseteq U$. This is possible because the $B_n$ form a base for $X$ and $X$ is regular (like in the 1-1-ness proof), these fact guarantee us the existince of enough pairs and thus enough functions defined for these pairs: we have some index $N$ coresponding to this $(n,m)$ pair (this is the re-indexing mentioned in the beginning) and we know that $f_N = g_{n,m}$ which was chosen in the beginning to obey $g_{n,m}[\overline{B_n}] = \{1\}$ and $g_{n,m}[X \setminus B_n] = \{0\}$, so as $x_0 \in B_n$ in particular $1=g_{n,m}(x_0) = f_N(x_0)$, and so $\pi_N(F(x))= f_N(x) = 1>0$ so that later in the proof we can conclude that $z_0 = F(x_0) \in \pi_N^{-1}[(0,\infty)]=V$, and so indeed $z_0 \in V \cap Z$
This is just by "construction" (how the $f_N = g_{n,m}$ were chosen) and the definitions of inverse images and projections and $F$ itself.
The proof (Munkres' I believe) is fine, really, but could do with a bit more explicitness for people not used to arguments like this with the product map. In my class at the time we first proved a more general embedding theorem for products and deduced the Urysohn metrisation theorem from that; all we need is that a second countable regular $T_1$ space is normal, and this implies (with the countable base again) that $X$ has a countable family of continuous $[0,1]$-valued functions that separates points, and points and closed sets. The embeddability of $X$ into $[0,1]^\omega$ follows then by the general theorem.
In the posted proof, these ideas are intermixed. IMHO it might be better to separate the embedding idea in general from the specific details for a regular second countable space. I've always found it clarifying (and the embedding theorem I mentioned also has other applications, so it makes sense to specify it separately. Munkres does that later in an exercise.