Can anyone give me an example of a Topological Vector Space that is not metrizable? I know that the neighborhood base of $0$ needs to be incountable, and all I can construct then is no topological vector space because the algebraic operations (especially multiplication) aren't continous.
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possible duplicate of Prove that some topology is not metrizable – Tomasz Kania Oct 15 '14 at 07:47
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@TomekKania I don't think it's a duplicate. There's a difference between proving that some given topology is not metrizable, and coming up with the non-metrizable topology yourself. – Najib Idrissi Oct 15 '14 at 09:08
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Take for instance $\mathbb{R}^{\mathbb{R}}$ with the product topology or the weak topology on an infinite-dimensional Banach space. You will find more examples here.
Tomasz Kania
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2The product topology on $\Bbb{R}^\Bbb{N}$ is metrizable, in fact the topology is polish. You will have to take uncountable products. – PhoemueX Oct 15 '14 at 07:47
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product topology on $R^n$ sounds good. if possible i'd like to stick to a finite dimensional. so the product topology there isn't the same as the topology induced by the common norms? – Bananach Oct 15 '14 at 07:47
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oh, okay. that's what I suspeceted. thanks for the examples and the link – Bananach Oct 15 '14 at 07:48
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@Bananach: On finite dimensional vector spaces, the product topology is just the "usual" topology, so you will have to take an infinite dimensional example. – PhoemueX Oct 15 '14 at 07:48
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@PhoemueX, you need the space to be Hausdorff as seen here. Otherwise any vector space equipped with indiscrete topology is not metrizable. – ChesterX Oct 15 '14 at 07:57
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@ChesterX: Sure, thanks :) What I tried to get at was that the product topology of the usual topology on the factors $\Bbb{R}$ is the same as the usual topology on $\Bbb{R}^n$. The general statement is that
every Hausdorff topology on a vector space which makes the vector space operations continuous is the usual topology.
I also ommited the addition concerning the compatibility with the vector space structure before, for simplicity.
– PhoemueX Oct 15 '14 at 08:00 -
my question is fully answered by your examples and terry's theorem 2, thanks for the quick help everyone – Bananach Oct 15 '14 at 08:05