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It is instructed in Munkre's topology that proof of imbedding theorem is almost the copy of step 1 of this post.It is instructed to just replace $n$ by $alpha$ and $mathbb R^{omega}$ by $mathbb R^J$.

Imbedding theorem:Let X be a space in which one point sets are closed.Suppose that {$f_{alpha}$}_{$alphain J$}} is an indexed
family of continuous functions $f_{alpha}:Xin mathbb R$ satisfying
the requirement that for each point $x_0$ of X and each nbhd U of
$x_0,$ there is an index $alpha$ such that $f_{alpha}$ is positive
at $x_0$ and vanish outside U.then the funcion $F:Xrightarrow mathbb
R^J$ defined by $F(x)=(f_{alpha})_{alphain J}$ is an imbedding of X
in $mathbb R^J.$if $f_{alpha}$ maps X into $[0,1]$ for each
$alpha,$ then $F$ imbeds $X$ in $[0,1]^J$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 an indexed collection of continuous functions $f_{alpha}:Xrightarrow [0,1]$ having the property that given any point $x_0$ of $X$ and any neighborhood $U$ of $x_0$,there exists an index $alpha$ such that $f_{alpha}(x_0)>0$ and $f_alpha$ vanishes outside $U$.

Proof:

Let {$B_{alpha}$} be the countable basis for $X$.For each pair $alpha,beta$ of indices for which $overline{B_{alpha}} subset B_{beta},$apply Urysohn's lemma to choose a continuous function $g_{alpha,beta}:Xrightarrow[0,1] $ such that $g_{alpha,beta}[overline{B_{alpha}}]={1}$ and $g_{alpha,beta}[X-B_{beta}]={1}$. Then the collection {$g_{alpha,beta}$} 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 $B_{alpha}$ so that $x_0 in B_{alpha}$ and $overline{B_{alpha}} subset B_{beta}$.Then $alpha,beta$ is a pair of indices for which function $g_{alpha,beta}$ is defined; and it is positive at $x_0$ and vanishes outside of $U$. Because the collection {$g_{alpha,beta}$} is indexed with a subset of indexing set $J$, gives us the desired collection {$f_{alpha}$}.

Given a function $f_{alpha}$ of step 1,take $mathbb{R}^{J} $ in the product topology and define a map $F: Xrightarrow mathbb{R}^{J}$ by the rule-
$F(x)=(f_{alpha})_{alphain J}$

We assert that $F$ is an embedding-

$F$ is continuous

Because $mathbb{R}^{omega}$ has a product topology and each $f_{alpha}$ is continuous.

$F$ is injective

Let $x,yin X$ such that $xneq yimplies$ there is some basic element $B_{alpha}$ that contains $x$ and misses y.By applying Urysohn's lemma there exists a function $f_{delta}:Xrightarrow [0,1]$ such that $f_{delta}(x)=1,f_{delta}(y)=0$.So,$F(X)neq F(y)$(because the image differs atleast in one coordinate)

$F$ is a homeomorphism of $X$ onto its image, the subspace $Z=F[X]$ of $mathbb{R}^{J} $.

Because each component of $F$ is continuous,so $F$ is continuous.**

$F:Xrightarrow F[X]$ 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 $z_0in F(U).$

We shall find an open set $W$ of Z such that $z_0in Wsubset F(U)$.

Let $x_0in U$ such that $F(x_0)=z_0.$Choose an index $theta$ for which $f_{theta}(x_0)>0$
anf $f_{theta}(X-U)=${$0$.}

Take the open ray $(0,infty) $ in $mathbb R$ and let $V$ be the open set $V=pi_{theta}^{-1}((0,infty))$ of $mathbb R^{J}$.

Let $W=Vcap Z;$ then $W$ is open in $Z$ by definition of subspace topology.We assert that $z_0in Wsubset F(U).$

First $z_0in W$ because $pi_{theta}(z_0)=pi_{theta}F(z_0)=f_{theta}(x_0)>0.$

Second,$Wsubset F(U).$For if $zin W$ then $z=F(x)$ for some $xin X$ and $pi_N(z)in (0,infty)$.

Since,$pi_{theta}(z)=pi_{theta}(F(x))=f_{theta}(x),$ and $f_{theta}$ 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 $mathbb R^J.$

Also,examine this proof critically,and if there is some scope of improvement please let me know...

You don't have a countable base for $X$. So any part of the proof that refers to those base elements must be scratched.

You just know 2 things about $X$:

• One-point sets in $X$ are closed. ($X$ is $T_1$)




• There is a family of continuous functions ${f_alpha: X to mathbb{R}: alpha in J }$ that obeys:

$(ast)$ For every $x_0 in X$ and every open neighbourhood $U$ of $x_0$, there exists some $alpha_0 in J$ such that $f_{alpha_0}(x_0) > 0 $ and $f_{alpha_0}(x)=0$ for all $x notin U$.

We then define $F:X to mathbb{R}^J$ by $F(x)= (f_alpha(x))_{alpha in J}$, or equivalently $pi_alpha circ F = f_alpha$ for all $alpha in J$. It is immediate that $F$ is continuous as the its compositions with all projections are continuous (universal property of maps into products, that Munkres has in its text).

So $F:X to Z=f[X]$ is also continuous, and onto by definition.

That leaves to check that $F$ is 1-1 and $F$ is open as a map onto $Z$.

If $x neq y$ we use the first property and take $U = Xsetminus {y}$ which is
an open neighbourhood of $x$. We apply $(ast)$ to get the promised $f_{alpha_0} : X to mathbb{R}$ with $f_{alpha_0}(x)> 0$ and $f_{alpha_0}(y)=0$.
So $F(x)$ and $F(y)$ differ at the $alpha_0$-th coordinate and so $F(x) neq F(y)$ and $F$ is 1-1.

With $x_0in U$, $U$ open in $X$, such that $F(x_0) = z_0 in F[U]$ a you chose,
we find by $(ast)$ (Applied to $x_0$ and $U$) again some $beta in J$ (you have $theta$ but I prefer to use $beta$ after $alpha$) such that

$$f_beta(x_0) >0 text{ and } f_beta(x) =0 text{ for } x notin Utag{1}$$

Then of course $$pi_beta(F(x_0)) = f_beta(x_0) in (0,infty)$$ so that $$z_0= F(x_0) in pi_{beta}^{-1}[(0,infty)] cap Z = W$$

The fact that $W subseteq F[U]$ is clear: $y in W$ implies $y in Z =f[X]$, so there is some $x in X$ such that $f(x) =y$.

If $x notin U$ we know that $f_beta(x) = 0$, by how we chose $beta$ in $(1)$, and so $F(x) notin pi_beta^{-1}[(0,infty)] =V$, but this cannot be as $F(x) in W= V cap Z$. So $x in U$ must hold and so $y=F(x) in F[U]$ showing the inclusion $W subseteq F[U]$.