# Please check the proof of Imbedding theorem.

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 {$$f_{\alpha}$$}_{$$\alpha\in J$$}} is an indexed family of continuous functions $$f_{\alpha}:X\in \mathbb R$$ 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 $$F:X\rightarrow \mathbb R^J$$ defined by $$F(x)=(f_{\alpha})_{\alpha\in J}$$ 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 $$f_{\alpha}:X\rightarrow [0,1]$$ having the property that given any point $$x_0$$ of $$X$$ and any neighborhood $$U$$ of $$x_0$$,there exists an index $$\alpha$$ 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 $$g_{\alpha,\beta}:X\rightarrow[0,1]$$ 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 $$F: X\rightarrow \mathbb{R}^{J}$$ by the rule- $$F(x)=(f_{\alpha})_{\alpha\in J}$$

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 $$x,y\in X$$ such that $$x\neq y\implies$$ there is some basic element $$B_{\alpha}$$ that contains $$x$$ and misses y.By applying Urysohn's lemma there exists a function $$f_{\delta}:X\rightarrow [0,1]$$ 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.**

$$F:X\rightarrow F[X]$$ 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 $$\theta$$ 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 $$W=V\cap Z;$$ 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 $$\pi_{\theta}(z_0)=\pi_{\theta}F(z_0)=f_{\theta}(x_0)>0.$$

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,$$\pi_{\theta}(z)=\pi_{\theta}(F(x))=f_{\theta}(x),$$ 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...

• We don't need Urysohn's lemma but just the property for the functions and the fact one point set aree closed. See my proof below. Bases are irrelevant. – Henno Brandsma Nov 24 '18 at 9:41

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 $$U = X\setminus \{y\}$$ 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 $$x_0\in U$$, $$U$$ 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

$$f_\beta(x_0) >0 \text{ and } f_\beta(x) =0 \text{ for } x \notin U\tag{1}$$

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 $$W \subseteq F[U]$$.

• :While showing injectivity I think ,by mistake you wrote $f_{\alpha_0}(y)=1$ instead of $f_{\alpha_0}(y)=0$. – P.Styles Dec 2 '18 at 13:12
• @P.Styles already edited. – Henno Brandsma Dec 2 '18 at 13:13