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In Rudin's Functional Analysis, Theorem 1.21 states

If $n$ is a positive integer and $Y$ is an $n$-dimensional subspace of a complex topological vector space $X$, then

(a) every isomorphism of $\mathbb{C}^n$ onto $Y$ is a homeomorphism, and

(b) $Y$ is closed.

In particular this implies that every $n$-dimensional complex topological vector space is homeomorphic to $\mathbb{C}^n$.

Am I missing something here? I thought that there were some separability axioms that had to hold (see Every finite dimensional topological separated vector space is isomorphic and homeomorphic to $\mathbb{K}^{n}$. for example).

Caleb Jonker
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    If $Y$ is a subspace of a complex topological vector space $X$, then $X$ must be Hausdorff (by definition). So Rudin is stating that every complex topological vector space of dimension $n$ is homeomorphic to $\mathbb{C}^n$. – Siminore May 02 '18 at 12:40
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    I don't own Rudin's book. For many authors, a topological group is by definition Hausdorff. – hunter May 02 '18 at 12:40
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    It's usual to insist that a TVS be Hausdorff. Does Rudin? – Angina Seng May 02 '18 at 12:40
  • Yeah, Hausdorff-ness for a TVS is equivalent to the identity giving a one-point closed set, which is a pretty weak/harmless assumption to impose. – Randall May 02 '18 at 13:06
  • Oh yes of course. Rudin assumes every point is a closed set and then proves Hausdorff as a result. – Caleb Jonker May 02 '18 at 13:14

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Am I missing something here? I thought that there were some separability axioms that had to hold.

In Rudin's definition a topological vector space is always T1, i.e. one-point sets are closed, this implies that $X$ is hausdorff (as proved there). Thus, $X$ with the indiscret topology is not a TVS according to Rudin's definition, because the indiscret topology is not hausdorff.

Note also that we always have a fixed topology on the ground field $\mathbb{C}$ without mentioning it. As usual, $\mathbb{C}$ is always equipped with the standard topology induced by the Euclidean norm. Indirectly, this implies in the finite-dimensional case the separability of $X$.

If $X$ is a complex topological vector space , then $\mathbb{C} \times X \rightarrow X$ via $(\alpha,x) \mapsto \alpha x$ is continuous in the product topology on $\mathbb{C} \times X$. In particular, for some fixed $x \in X$, $x \neq 0$ (we can assume $X \neq 0$, otherwise we don't have to show anything), the map $f \colon \mathbb{C} \rightarrow X$ via $f(\alpha) = \alpha x$ is continuous.

For example, this forces that the topology of $X$ cannot be discret! In this case this map is not continuous, because $f^{(-1)}(\{x\})$ is not open in $\mathbb{C}$. So $X$ is a topological vector space with the discret topology only if $\mathbb{C}$ also carries the discret topology. As already said, implicitly we always consider the Euclidean topology on $\mathbb{C}$.

p4sch
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