I am trying to show that in any finite dimensional normed vector space, the unit ball is compact. To do this, I first proved that:
$(X,|\cdot |_X)$ normed vector space is: locally compact $\iff$ the ball $\widetilde{B}(0;1)=\{x\in X \;| \;|x|_X\leq 1\}$ is compact.
having done this, I would need to prove that:
$(X,\tau)$ topological vector space (Hausdorff) is: locally compact $\iff$ of finite dimension.
To do this, I found this document: https://terrytao.wordpress.com/2011/05/24/locally-compact-topological-vector-spaces/ where "Theorem 1" represents the implication "$\Rightarrow$" of the statement that I want to prove. There are some steps that unfortunately I have not been able to understand, in particular the demonstration begins with:
Let V be a locally compact Hausdorff space, thus there exists a compact neighbourhood K of the origin. Then the dilate $\frac{1}{2}K$ is also a neighbourhood of the origin, and so by compactness K can be covered by finitely many translates of $\frac{1}{2}K$...
I don't understand why we are sure that $\frac{1}{2}K=\{\frac{k}{2}\}_{k\in K}$ is a neighborhood of the origin. To affirm this we should be certain that in the topology $ \tau $ there is also an open whose closure is equal to $\frac{1}{2}K$.
No?