Talk:Kolmogorov's normability criterion

Why T1 ?
I was wondering why one uses the T1 axiom here. It is known that for a topological vector space (in fact for every topological group) T1 is equivalent to the seemingly weaker T0 and to the (seemingly stronger) Hausdorff axiom. So, I was expecting to see either the weakest notion T0 so that the statement looks the strongest, or to see Hausdorff which is the most well-known of all separation axioms ... --2003:C9:7F1C:4958:1462:88B4:5AB5:B091 (talk) 09:01, 25 May 2022 (UTC)