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Let $Y$ be a compact, separable metric space and $X=C(Y)$ Banach space. There are many criteria for a linear subspace $Z\subseteq X$ to be dense; notably the Stone-Weierstraß theorem.

Are there theorems giving conditions on non-linear subsets $Z$ of $X$ to be ε-dense? Particularly, if I know that

  • $Z$ comprises "partial sums" of at-most $N$ elements; i.e., $$ Z=\left\{ k+\sum_{i=0}^N k_iy_i:\, y_i \in \tilde{Y}, k,k_i \in \mathbb{R} \right\}, $$ for some positive integer $N>0$ and some subset $\tilde{Y}\subseteq Z\subseteq C(Y)$,
  • $\operatorname{span}(Z)$ is dense in $X$?
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    $\begingroup$ Do you mean that Z is comprised of "partial sums"? $\endgroup$
    – Ruy
    Nov 25, 2019 at 12:41
  • $\begingroup$ Yes, thank you. $\endgroup$
    – ABIM
    Nov 25, 2019 at 13:10

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If your set $Z$ is $\epsilon$-dense, then it is also dense! This is in fact so for every normed space $X$, and for every subset $Z\subseteq X$ which is positively homogeneous in the sense that $\alpha Z = Z$, for every scalar $\alpha>0$. The reason is as follows, for any point $x$ in $X$, and any $\alpha>0$, one has that $$ d(\alpha x,Z) = d(\alpha x,\alpha Z) = \alpha \,d(x,Z), $$ where $d$ is the distance function.

Thus, if $Z$ is $\epsilon$-dense, one has for every $x$ in $X$, and every natural number $n>0$, that $$ d(x,Z) = (1/n)\,d(nx,Z) \leq (1/n)\varepsilon, $$ from where you deduce that $d(x,Z)=0$

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  • $\begingroup$ Nice! This was subtle but very interesting! Thank you. $\endgroup$
    – ABIM
    Nov 29, 2019 at 9:03

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