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Every topological vector space is completely regular. My question is, is the converse true? That is, is every completely regular topology induced by some topological vector space?

If not, does anyone know of a counterexample?

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  • $\begingroup$ isn't every metric space completely regular? in which case, $\{0,1\}$...... $\endgroup$ – Dan Rust Feb 19 at 1:55
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    $\begingroup$ So there are obvious counterexamples to the statement "If $X$ is completely regular, then $X$ is a topological vector space". Do you perhaps want a proof/contradiction to the statement "If $X$ is completely regular, then $X$ embeds in a topological vector space"? $\endgroup$ – Aweygan Feb 19 at 1:59
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    $\begingroup$ @DanRust Well, $\{0,1\}$ is a ($1$-dimensional) topological vector space over the (discrete) topological field $\mathbb{F}_2$. But a discrete space with $6$ points does the trick, since any finite vector space has prime power order. $\endgroup$ – Alex Kruckman Feb 19 at 2:01
  • $\begingroup$ @AlexKruckman good point. I was just thinking about real/complex coefficients. I think Aweygan's question is maybe more what OP was aiming for though. $\endgroup$ – Dan Rust Feb 19 at 2:03
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    $\begingroup$ Just in case the question was intended to be as @Aweygan suggested: Every completely regular space embeds into a power of the interval $[0,1]$ with the product topology. (This is the main step in one construction of the Stone-Cech compactification.) So it embeds into a power of $\mathbb R$, which is a topological vector space. $\endgroup$ – Andreas Blass Feb 19 at 3:12
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Just in case the question was intended to be as @Aweygan suggested: Every completely regular space embeds into a power of the interval [0,1] with the product topology. (This is the main step in one construction of the Stone-Cech compactification.) So it embeds into a power of ℝ, which is a topological vector space.

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