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?



$\Sigma_*$-product is not $\sigma$-countably compact
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 $K\_n$.


