By "weakly separable" I mean the notion for uniform spaces used by David Wigner and Lawrence Brown: a uniform space is weakly separable if any uniform cover has a countable subcover. For a topological group $G$ this means for any open set $U$ in $G$, the cover $\{gU\}_{g \in G}$ has a countable subcover, or equivalently every cover $\{Ug\}_{g \in G}$ has a countable subcover. Thank you.
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Let's take the group $G = \mathbb Z^T$, where $T$ is an uncountable set.
Is $G$ weakly separable? Yes: If $U$ is any neighborhood of $0$ in $G$, then there exists a finite set $T_0 \subset T$ so that $$ U \supseteq V := \{\phi \in G \colon \phi(t)=0 \text{ for all } t \in T_0\} $$ Countably many translates of $V$ cover $G$: namely translates by all $\psi \in G$ supported by $T_0$.
If $G$ Lindelof? No. Steen & Seebach, Example 103.
answered Feb 24, 2014 at 17:33
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$\begingroup$ Thank you! I'm not sure what $\delta$ has to do with it though.... $\endgroup$ Commented Feb 25, 2014 at 14:14
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1$\begingroup$ OK $\delta$ can be removed. My first version was $\mathbb R^T$, and $\delta$ was used for the neighborhood. $\endgroup$ Commented Feb 25, 2014 at 15:29