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I would really appreciate some comments on my understanding of the relation between complete function spaces and dense subsets.

First some definitions.

  • Let $X$ be a complete scalar field, and $k$ a function $k: X \times X \mapsto \mathbb{R}$.
  • Let $span(k)$ be the space of functions defined in the following way, \begin{equation} span(k) = \{\sum_{i=1}^{D} \alpha_{i} k(x_{i},\cdot) \mid D \in \mathbb{N}, D < \infty, \{\alpha_{i}\}_{i=1..D} \subset \mathbb{R}^{D},\{x_{i}\}_{i=1..D} \subset X^{D} \}. \end{equation}
  • Let $H(k)$ be a complete Hilbert space of functions such that $span(k)$ is a dense subset of $H$.

My questions

  • Since every finite dimensional vector space over a complete field is complete; am I right in concluding that if $span(k)$ is finite dimensional, then $H(k) = span(k)$ ?
  • And conversely, if $H(k)$ is finite dimensional, then $H(k) = span(k)$ ?
  • I am trying to understand this in relation to the fact that it is (under certain conditions) possible for a vector space to have several dense but disjoint subsets; the rationals and irrationals in $\mathbb{R}$ is an example. If this holds also for a finite dimensional $H(k)$, then obviously, it does not hold that $H(k)= span(k)$. When is it possible to find several disjoint dense subsets of $H(k)$?

I must have misunderstood something along the way, and would really appreciate any help in clarifying!

Note: the above questions are relevant to Reproducing kernel Hilbert spaces, where $k$ is the "kernel", and $H(k)$ is the RKHS.

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In a finite dimensional Hilbert space all subspaces are closed. Therefore the only dense subspace is the whole space. Therefore your first two points are correct.

As for the last one, the real and the irrationals are both dense but the irrationals are not a subspace. For subspaces, my favorite example are the subspace of polynomials and the subspace of $\textit{trigonometric}$ polynomials. These are two distinct subspaces that are both dense in say $L^2[0,1]$.