Let $X$ be a compact space. Show that the following statements are equivalent:
a) $X$ is homeomorphic to a compact subset of $\mathbb{R}^n$
b) There are functions $f_1,\dotso, f_n\in C(X)=\{f:X\to\mathbb{C}\quad\text{continuous}\}$, such that every linear combination of functions from the form
$x\mapsto f_1(x)^{k_1}\cdot\dotso\cdot f_n(x)^{k_n}$ with $k_1,\dotso, k_n\in\mathbb{N}_0$ are dense in $C(X)$.
Hello,
I want to proof the implication a) "$\Rightarrow$" b). For this task the hint is given, that we need the theorem of Stone-Weierstrass and Urysohn's Lemma.
To proof the first implication, I need a subalgebra $\mathcal{A}\subseteq C(X)$, which contains every linear combination of functions from the form above:
$x\mapsto f_1(x)^{k_1}\cdot\dotso\cdot f_n(x)^{k_n}$ with $k_1,\dotso, k_n\in\mathbb{N}_0$
How can I note this set $\mathcal{A}$?
$\mathcal{A}:=\{\sum_{i=1}^k f_{i_1}^{k_{i_1}}\cdot\dotso\cdot f_{i_n}^{k_{i_n}}, f_{i_1},\dotso, f_{i_n}\in C(X), k_{i_1},\dotso, k_{i_n}\in\mathbb{N}_0\}$
I am not sure, because the product of these functions can contain any function, and I do not really know, how to note it as a set.
After that I have to prove that $\mathcal{A}$ is a subalgebra and holds the three conditions of Stone-Weierstrass:
1) contains the constant functions
2) self-adjoint
3) seperates points
Thanks in advance for any help.
For $a \implies b$, consider a subset $X \subseteq \Bbb R^n$, and the collection of functions $$ f_1(\mathbf x) = x_1, f_2(\mathbf x) = x_2,\dots, f_n(\mathbf x) = x_{n} $$ Note that $\mathcal A$ contains the constant function $$ g(x) = (f_1)^0(f_2)^0 \cdots (f_n)^0 $$ moreover, $\mathcal A$ separates points since for any two elements $\mathbf a = (a_1,\dots,a_n) \in X, \mathbf b = (b_1,\dots, b_n)\in X$, we can take the function $$ g(\mathbf x) = (x_1-a_1)^2 + (x_2-a_2)^2 + \cdots + (x_n-a_n)^2 $$ which satisfies $g(\mathbf a) = 0$ and $g(\mathbf b) = \|\mathbf a - \mathbf b\|^2 > 0$.
If $X$ is homeomorphic to a subset of $\Bbb R^n$, all of this still works. In particular, suppose that $h : X \to Y \subset \Bbb R^n$ is a homeomorphism. Then, for $x \in X$, we can define $$ \tilde f_i(x) = f_i (h(x)) $$ where $f_i$ are the "projections" as defined above.
For $b \implies a$, I think the idea is to define the map $h:X \to \Bbb R^n$ by $$ h(x) = (f_1(x),\dots,f_n(x)) $$ and show that this map is a homeomorphism. However, in the problem as written, the $f_i$ are complex-valued, which means that this isn't even a map into $\Bbb R^n$.