So, this question came up in a discussion today and I thought I'd post it here. Given that the ring of symmetric functions $\Lambda$ can be equipped with the Hall scalar product. Is it also a complete metric space?
Edit: Supposing that it is false I thought an amusing counter example might be using the sequence
\begin{equation}(e_{0}, e_{0}+e_{1},e_{0}+e_{1}+e_{2},\dots)\end{equation}
As the limit would result in Macdonalds classic example of an element of $\hat{\Lambda}$ with
\begin{equation}\prod_{i \geq 1}(1 + x_{i}) = \sum_{i \geq 0}e_{i}.\end{equation}
Of course this sequence would need to be Cauchy and there lies the problem. If we allow rational coefficients rather than just integer coefficients we could instead consider elements of the form
\begin{equation}\sum_{i=0}^{j}\frac{e_{i}}{(i+1)^{2}} \Rightarrow (e_{0}, \frac{1}{4}(e_{0}+e_{1}), \frac{1}{9}(e_{0},e_{1},e_{2}),\dots).\end{equation}
Here the limit is indeed going to have unbounded degree so we have a counterexample for $\Lambda_{\mathbb{Q}}$. However, for sticking with the more classical integer coefficients. I think this requires all the Cauchy sequences to be sequences that stabilise. Which would mean that it is indeed complete, but the only Cauchy sequences are rather uninteresting.
Edit 2: Obviously not a Hilbert space, my definition was bad.
I don't really understand the question. If you consider only symmetric functions over $\mathbb{Z}$ or $\mathbb{Q}$ you don't even have a vector space over $\mathbb{R}$ so you certainly don't have a Hilbert space. If you consider symmetric functions over $\mathbb{R}$ then symmetric functions over $\mathbb{R}$ are countable-dimensional, and infinite-dimensional Hilbert spaces must be uncountable-dimensional. Explicitly, the Schur polynomials $s_{\lambda}$ form an orthonormal basis with respect to the Hall inner product (in the Hamel, not the Hilbert space, sense) but no nontrivial infinite sum $\sum c_{\lambda} s_{\lambda}$ converges, even the ones satisfying $\sum |c_{\lambda}|^2 < \infty$.