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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 {$f_{\alpha}$}_{$\alpha\in J$}} 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...

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  • $\begingroup$ 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. $\endgroup$ – Henno Brandsma Nov 24 '18 at 9:41
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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 $W \subseteq F[U]$.

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

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