I am interested in a reference for the following fact (or a similar result).
PROPOSITION. Let $X$ denote the set of real null sequences; i.e., the set of $(a_n)_{n=0}^{\infty}$ with $a_n\to 0$, with the usual (product) topology. Then $X$ is not $\sigma$-compact.
By definition, $X$ is $F_{\sigma \delta}$ in the space of all real sequences. Since $X$ is a subset of the space of all bounded real sequences, which is $\sigma$-compact, the claim is equivalent to saying that $X$ is not $F_{\sigma}$. The proof of the proposition is not difficult (see below). I am using a similar proof in my recent preprint that shows that the escaping set of a transcendental entire function is never $\sigma$-compact. It would therefore be useful to know of a reference for the above, which feels as though it should be classical.
Proof of the Proposition. If $\underline{a} = (a_n)_{n=0}^{\infty}\in X$ is a null sequence, and $\varepsilon>0$, set $$ n_{\varepsilon}(\underline{a}) := \min\{n\geq 0\colon |a_n| < \varepsilon\} < \infty.$$ Clearly for all sequences $(\varepsilon_k)_{k=0}^{\infty}$ and $(N_k)_{k=0}^{\infty}$ with $\varepsilon_k\to 0$, there is $\underline{a}\in X$ such that $$ n_{\varepsilon_k}(\underline{a}) > N_k$$ for all $k\geq 0$. (This is just saying that there are sequences that tend to zero arbitrarily slowly.) If $A\subset X$ is compact, then $$ n_{\varepsilon}(A) := \max_{\underline{a}\in A} n_{\varepsilon}(\underline{a}) < \infty.$$ Let $(A_k)_{k=0}^{\infty}$ be a sequence of compact subsets of $X$. Let $\underline{a}\in X$ be such that $$ n_{1/k}(\underline{a}) > n_{1/k}(A_k)$$ for all $k\geq 0$. Then $\underline{a}\notin A_k$ for all $k$, and hence $$ X \neq \bigcup_{k=0}^{\infty} A_k,$$ as claimed.
