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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?


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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. – Matt Feb 27 '12 at 17:32
@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). – Tim Feb 27 '12 at 17:59
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 $T_0$). – Brian M. Scott Feb 28 '12 at 0:24

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