of $\sigma$-compact topological groups and for each $i$ let $e_i$ be the identity element of $X_i$. Then the subgroup $$G:=\{f\in\prod_{i\in I}X_i:\vert\{j\in I:f(j)\neq e_j\}\vert< \aleph_0\}$$ is $\sigma$-compact.

\[corsoncor\] For each infinite cardinal number $\kappa$ there is a ${\sf T}_0$ $\sigma$-compact Rothberger bounded group of cardinality $\kappa$.

Let cardinal number $\kappa$ be given, and take $I$ to be $\kappa$. For each $i\in I$ take $X_i$ to be ${{\mathbb Z}}$, the additive group of integers. Now consider the group $G$ as in Corson’s Theorem. $G$ is in fact Rothberger bounded. To see this let for each $n$ a neighborhood $U_n$ of the identity element of $G$ be given. We may assume each $U_n$ is a basic open set, and thus that there is a finite set $F_n\subseteq I$ and for each $i\in F_n$ a neighborhood $N_{n,i}$ of $e_i$ such that $U_n$ is of the form $\{f\in G:(\forall i\in F_n)(f(i)\in N_{n,i})\}$. Now $C = \cup_{n<\omega}F_n$ is a countable subset of $I$ and $G_C = \{f\lceil_C:f \in G\}$ is evidently Rothberger bounded in $\prod_{i\in C}X_i$. For each $n$ choose a $g_n\in G_C$ such that $G_C\subseteq \cup_{n<\omega}g_n*U_n\lceil_C$. For each $n$ choose $f_n\in G$ with $f_n\lceil_C = g_n$. Then it follows that $G\subseteq \cup_{n<\omega}f_n*U_n$. Thus $G$ is 