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First of all I must apologize for the vague title and am open to suggestions.

This is not a Homework Assignment but something I once again encountered while reading a very compactly written paper.

$\tilde{\mathcal{C}}([0,1]^2,\mathbb{R}))$ is defined as follows for $f\in \tilde{\mathcal{C}}([0,1]^2,\mathbb{R}))$ the following holds:

  1. $f$ is continuous
  2. $f(x)=0 \quad \forall x\in [0,1]^2\setminus [0,1)^2$

I am looking for a countable family of continuous functions $\mathcal{F}:=(f_m)_{m\in\mathbb{N} }$ on $[0,1]^2$ so that the following requirements are satisfied

  1. $f\in \mathcal{F}\implies f(x)=0 \quad \forall x\in [0,1]^2\setminus [0,1)^2$
  2. the closure of the linear span of $(f_m)_{m\in\mathbb{N}}$ consists of all continuous Functions on $[0,1]^2$

According to the authors such a sequence is supposed to exist. But I do not know what the $f_m$ would like or why the family with the aforementioned properties exists. So I am looking for an explicit construction of the $f_m$ or a theorem which proofs the existence of such a sequence.

My first guess was $f_{n.m}(x_1,x_2):=x_1^nx_2^m$ for with Stone-Weierstrass it fullfills the second requirement. Unfortunately it does not meet the first one and I am somewhat at a loss.

Has anybody encountered a similar construction or knows of a theorem that might help?

As always thanks in advance :)

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    @ Davide: yes this is actually what i wanted ^^2012-07-30

1 Answers 1

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We have that $\widetilde{\mathcal C}$ is a (closed) subspace of $\mathcal C[0,1]^2$. The latter is separable, and so will be $\widetilde{\mathcal C}([0,1]^2,\Bbb R)$. So a countable dense family $\mathcal F$ will do the job.