$\Sigma_*$-product is not $\sigma$-countably compact In Arhangel'skii's book "Topological function spaces" there is a part where the author uses that, if $\kappa>\omega$ is a cardinal number, then the space $$\Sigma_*(\kappa):=\left\{x\in \mathbb{R}^{\kappa} : \forall \varepsilon>0\left(\left|\left\{\alpha<\kappa : |x_\alpha|\geq \varepsilon\right\}\right|<\omega\right)\right\}$$ is not $\sigma$-countably compact. Although proving that the space $\Sigma_*(\kappa)$ is not countably compact is not hard, I have not been able to see why it is not $\sigma$-countably compact. The only promising idea I had was to try to see that $\Sigma_*(\kappa)$ has uncountable extent, but I haven't been able to find an uncountable subspace without accumulation points either.
Is it possible to argue that $\Sigma_*(\kappa)$ does indeed have uncountable extent, or must it be shown that $\Sigma_*(\kappa)$ is not $\sigma$-countably compact in another way?
 A: Let $X =\Sigma_*(\omega) =  \{ f \in {\mathbb{R}}^\omega : \forall \epsilon > 0,  \{ x \in \omega : |f(x)| < \epsilon\} \mbox{ is finite.} \}  $.  Since $X$ is homeomorphic to a closed subset of  $\Sigma_*(\kappa)$ and is metrizable, it is enough to show that $X$ is not $\sigma$-compact.
[The idea is this: Suppose $X = \cup_{n \in \omega}K_n$ where each $K_n$ is compact.  Partition $\omega$ into infinitely many infinite sets $S_n, n \in \omega$.  Define a function $G \in X$ which fails to be in $K_n$ for any $n$ because the restriction of $G$ to $S_n$ is a function which is not the restriction to $S_n$ of any element of $K_n$.]
It is convenient to  make a general construction.  Let $S = \{s_1, s_2, ...\}$, where $s_1 < s_2 < ...$, let $K$ be a compact subset of $X$, and let $r$ be a non-zero real number. For $m = 1, 2, ...$ let $f_m^{S,K,r} \colon S \to \{0, r\}$  be the function given by $f_m^{S,K,r}(s_k) = r$ if and only if $k \leq m$.  Also, let $ f_0^{S,K,r} \colon S \to \{0, r\}$ be the zero function.
For $m =   1, 2, ...$ let $A_m^{S,K,r} = \{f \in K : f \restriction S = f_m^{S,K,r}\}.$ $A_m^{S,K,r}$ might be empty for every $m$  (which can happen, for one example, if $h(s_1) \ne r$ for every $h \in K$).  In this case, or if the restriction to $S$ of every element of $K$ is the zero function, let $m_0 = 0$. If there is a non-empty set $A_m^{S,K,r}$, there is a largest $m$ such that $A_m^{S,K,r} \ne \emptyset$.  The reason is that if there were not such a largest $m$, the compactness of $K$ would mean that $K$ contained an element whose restriction to $S$ is the function which is identically $r$, and $X$ does not contain such an element.  In this case let $m_0$ be the largest value of $m$ such that $A_m^{S,K,r} \ne \emptyset$.  Let $g^{S,K,r} = f_{m_0 + 1}^{S,K,r}$. Then $g^{S,K,r}$ has the property that it assumes the value $r$ only finitely often and it is not the restriction to $S$ of any element of $K$.
Let $K_1, K_2, ...$ be compact subsets of $X$ (whose union we want to show is not all of $X$). Let $\{S_1, S_2, ...\}$ be a partition of $\omega$ into infinitely many infinite sets.  Let $G \colon \omega \to {\mathbb R}$ be the function whose restriction to $S_n$ is $g^{S_n,K_n,\frac{1}{n}}$.  Then $G \in X$ because $G(x) \geq \frac{1}{n}$ only for $x \in S_1, \cup ... \cup S_n$ and on each set $S_k$, $G$ is non-zero only finitely often.  Finally, if $n$ is a positive integer, $G \not\in K_n$ because the restriction of $G$ to $S_n$ is a function which is not the restriction to $S_n$ of any element of $K_n$.
