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Definition

A topological vector space $V$ is a space equipped with a topology $\mathcal{T}$ for which the vector sum $+:V\times V\rightarrow V$ and scalar multiplication $*:\Bbb{K}\times V\rightarrow V$ are continuous.

Statement

Let be $V$ a topological vector space. If $\text{dim}V<\aleph_0$ then there exist a norm $\|\cdot\|$ on $V$ that induces the same topology that makes $V$ a topological vector space.

Unfortunately I can't prove the statement: so I ask to prove it and if it is generally false I ask if it is true for field of real numer $\Bbb{R}$ or in some others cases. Anyway I know that any normed space is a topological vector space. So could someone help me?

Antonio Maria Di Mauro
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    What about the discrete topology on $V$? – Lee Mosher May 12 '20 at 16:55
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    It feels like there is more information that is assumed but not stated in the question. For example, is it assumed that the underlying field is $\mathbb R$ or $\mathbb C$? Is there any assumption about how the topology is constructed? – User8128 May 12 '20 at 16:59
  • @RobertFurber I'm interested at the case where $V$ is a vector space on a general field $\Bbb{K}$ – Antonio Maria Di Mauro May 12 '20 at 17:10
  • @AntonioMariaDiMauro I don't have a complete proof ready to go, but my suspicion is that the continuity of addition and scalar multiplication should be sufficient to show that the space is Hausdorff (or T1, or whatever separation condition is really needed), from which Urysohn can be invoked. The argument is not particularly constructive, but I think it will get the job done. – Xander Henderson May 12 '20 at 17:19
  • @XanderHenderson Okay, but I don't understand what can I argue from what you say. So it seems that you claim that any topological vector space is hausdorff separable: and so? – Antonio Maria Di Mauro May 12 '20 at 17:22
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    @AntonioMariaDiMauro Urysohn's lemma asserts that if a topological space is sufficiently separated (I don't recall the precise requirement; my recollection of the proof is that it follows from separating closed sets---so perhaps normality is required?), then it is metrizable. By definition, if a space is metrizable, then it has a metric which gives the same topology. A metric $d$ on a vector space induces a norm via $|x| = d(x,0)$. The key point (to my argument) is applying Urysohn's lemma, which should be Google'able. – Xander Henderson May 12 '20 at 17:26
  • And, again, I don't know if this line of reasoning works. It is my off-the-cuff intuition for how to obtain the desired result. – Xander Henderson May 12 '20 at 17:28
  • Umm... it seems that your Urysohn's lemma is different from mine. So I know that Urysohn's lemma claim that if $X$ is a normal space and if $F$ and $G$ are closed and disjoint then there exist a continuous function $f:X\rightarrow [0,1]$ such that $f[F]={0}$ and $f[G]={1}$. – Antonio Maria Di Mauro May 12 '20 at 17:31
  • Yes, and this gives you the metric. Try Googling "Urysohn metrization theorem." – Xander Henderson May 12 '20 at 17:34
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    If the definition of a topological vector space that you use doesn't include the condition that the space is Hausdorff (that can be via requiring it is $T_1$ for example, or $T_0$; for uniform spaces these imply $T_2$), it is trivially wrong, for the topology induced by a norm is always Hausdorff. If the definition does include Hausdorffness, it is true for topological vector spaces over $\mathbb{R}$ or $\mathbb{C}$. If you look at other scalar fields, what are the conditions on the field? – Daniel Fischer May 12 '20 at 17:44
  • @DanielFischer I edited the question: read it. Anyway I have posted this question after I read the first answer of this question: perhaps this could help you. Could you help me, please? – Antonio Maria Di Mauro May 12 '20 at 17:49
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    So the definition does not imply Hausorffness, and we're in "trivially false" territory. Take a Vector space $V$ of dimension $n > 0$, an endow it with the indiscrete topology. Thus all maps from any topological space to $V$ are continuous, hence this makes $V$ a topological vector space. But as mentioned above, topologies induced by norms are Hausdorff, thus the indiscrete topology on a nontrivial vector space is never induced by a norm. – Daniel Fischer May 12 '20 at 17:54
  • @DanielFisher Okay, so what can you say about what is written in the first answer of the question that I indicated to you? – Antonio Maria Di Mauro May 12 '20 at 17:56
  • It tacitly assumes that the definition of a topological vector space includes Hausdorffness (that's quite common, though not universal), and it also tacitly assumes that the scalar field is $\mathbb{R}$ or $\mathbb{C}$ endowed with the standard topology (also a common assumption), or possibly some "sufficiently similar" field. For completely arbitrary scalar fields, the notion of a norm doesn't even make sense. – Daniel Fischer May 12 '20 at 18:01
  • @DanielFisher Okay. However reading the answer it seems to me that who answered don't use the hypotesis of topological vector space to make a norm: what do you say about this? – Antonio Maria Di Mauro May 12 '20 at 18:04
  • This: "To do that, we use the fact that a finite dimensional topological vector space can be equipped with a norm which gives the same topology (in fact it's the unique one)" uses the tacit assumptions. (And, I don't get pinged because you forgot the 'c' in my name. If you use tab-completion, you need not worry about mis-spelling user names.) – Daniel Fischer May 12 '20 at 18:27
  • I think that G. G. Gould defines a Hausdorff topological field $k$ that is not first-countable in Definition 2.3, the underlying field of which is $\mathbb{R}(x)$. Any Hausdorff topological field that is not first-countable is a non-normable (because non-metrizable) 1-dimensional Hausdorff topological vector space over itself. Unfortunately I don't have the leisure to verify the details, so this is a comment, not an answer. – Robert Furber May 12 '20 at 19:51

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