Closed set in a finite-dimensional normed space.

Question

A finite-dimensional normed space X is complete. Every complete space is also closed. If I take linearly independent $x_1,\dots,x_n \in X$, then the set $\{\sum_{i=1}^{n} a_i x_i : a_i > 0\}=:A$ is finite-dimensional. Hence this set should be closed. But obviously $0 \notin A$. But I can choose every scalar $a_i$ smaller and smaller hence I can obtain a sequence in $A$ converging to $0$. Therefore it shouldn't be closed in my opinion.

Answer 1

Your set $A$ is not a subspace. Actually, take $n=1$, $X=\mathbb R$, $x_1=1$, and now you are saying that $(0,\infty)$ should be closed. Why would that be?