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From the first two, we can say every topological vector space is completely regular, can't we?

Does a TVS need to be hausdorff in order to be completely regular?

Thanks!

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    I think this depends on your definitions. Some people require TVS to have the condition $T_0$, in which case they are automatically Hausdorff, so I assume you don't require this. Some people use the term completely regular to mean $T_{3\frac{1}{2}}$ in which case Hausdorff is needed. It seems if completely regular is the definition where you don't require Hausdorff, then it should be true.2012-02-27
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    @Matt: Thanks! (1) Could you point out what "completely regular" means in the first one and the third one? (2) Do "topological vector space" in the last two include $T_0$ in their definitions (Wiki also mentions this, but its definition doesn't assume $T_0$ in TVS definition).2012-02-27
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    In the first and second ones, *completely regular* does not include $T_0$; in the third it does (or might as well, since Hausdorff implies $T_0$).2012-02-28

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