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Let $V$ be a real vector with finite dimension, let's say $\operatorname{dim}_\mathbb{R}(V) = n$. The dual space of $V$ is defined by $V^* = \{ f : V \to \mathbb{R} \text{ linear} \}$.

I was told that one can define a topology on $V^*$, $\operatorname{GL}(V)$ and $\operatorname{GL}(V^*)$ but I have no idea how it is defined. Could someone tell me how exactly the open resp. closed set over this topology look like? Thanks a lot!

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All topologies induced by a norm over finite dimensional spaces (of the same dimension) are equivalent to the standard topology on $\mathbb R^n$ so it is very much up to you how to choose your topology, noting that $\dim V^*=\dim V$.

If you'd like, a dual vector in $V^*$ can be identified with some vector $v=(v_1, \dots v_n) \in V$ via the dual basis construction (not canonical.)

One can then define the usual open ball topology here induced by the norm $\|v\|:=\sqrt{v_1^2+ \dots+v_n^2}$.

A slightly more sophisticated approach, more amenable to generalization is the idea of the "continuous dual," which consists of functions that are continuous $f:V \to k$ (for finite dimensional spaces, this is every linear function.) Here, the topology would be induced by the norm $$\sup_{v \in V} \frac{\|f(v)\|}{\|v\|}$$ where $\| \cdot\|$ is any norm on $V$.

All in all: under the usual identification of a dual vector space with the original space, closed sets look exactly how you would expect them to.

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    $\begingroup$ What you say in your first sentence is not correct. For example, the topology induced by the discrete metric on $\mathbb R$ is surely not equivalent to the standard topology. You mean all topologies under which vector addition and scalar multiplication is continuous. $\endgroup$ – amsmath Oct 29 '17 at 19:17
  • $\begingroup$ Excuse me. Yes, you are correct $\endgroup$ – Andres Mejia Oct 29 '17 at 19:23

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