0

I found this question: Prove that if f in $C(X \times Y)$ then there exists functions.
And I was going to make a comment there but since is too old, I don't belive I'll get an answer.

The only answer in the question, says that you have to use Stone-Weierstrass, however I don't see how. The sentence that we use in class is:

Let $K$ be a compact metric space and let $\mathcal A$ be a subset of $C^0(K)$ with the following properties:
a) $\lambda\varphi+\mu\psi\in\mathcal A$ for any $\varphi,\psi\in\mathcal A$ and $\lambda,\mu\in\Bbb R$.
b) $\varphi\psi\in\mathcal A$ for any $\varphi,\psi\in\mathcal A$.
c) $1\in\mathcal A$.
d) if $x_1\neq x_2$ in $K$, then exists $\varphi\in\mathcal A$ such that $\varphi(x_1)\neq \varphi(x_2)$.
Then $\mathcal A$ is dense in $C^0(K)$, ie, given a countinous function $f:K \to \Bbb R$, there exists a sequences of funcionts $(\varphi_k)_k$ in $\mathcal A$ such that it converges uniformly to $f$ in $K$.

I think, in this case $K=X\times Y$, $\mathcal A=\{\varphi\in C^0(X\times Y):\varphi(x,y)\mapsto \sum_{i=1}^kf_i(x)g_i(y)\}$ and probably: $(\varphi_k)=(\sum_{i=1}^kf_i(x)g_i(y))_k$
... but now what? Conditions a), b), and c) don't look difficult to prove, however d), wich I think is the important one, is not so clear, if we have $(x_1,y_1),(x_2,y_2)\in X\times Y$ then don't we need to know more about $f_i,g_i$?

This isn't clear to me, we want this to work for any continous function $f:X\times Y \to \Bbb R$, isn't it? so that would mean that what we're trying to prove is that those $f$'s are the uniform limit of functions $f_1(x)g_1(y)+...+f_n(x)g_n(y)$ when $n\to\infty$. Maybe I'm just rambling with this, but I'm getting quite counfused.

Community
  • 1
Ana Galois
  • 2,466

1 Answers1

1

Now $\mathscr A$ is given by

$$\mathscr A = \{ \phi \in C^0(X\times Y):\ \phi(x, y) = \sum_{i=1}^k f_i(x) g_i(y),\ f\in C^0(X),\ g\in C^0(Y)\}$$

Let $(x_1, y_1)$ and $(x_2, y_2)\in X\times Y$. Then either $x_1 \neq x_2$ or $y_1 \neq y_2$. Assume the first case, let $f\in C^0(X)$ such that $f(x_1) = 1$ and $f(x_2) = 2$. Then $\phi(x, y) = f(x) \in \mathscr A$ would satisfy (d). Similar for $y_1\neq y_2$.