Is a set of positive linear combination of basis of a finite dimensional vector space is an open set?

Question

Is the following statement true?

Let $V$ be a vector space of dimension $n$ and let $\{x_1, \dots, x_n\}$ be a basis of $V$. Then, $$\left\{z=\sum_{i=1}^n \alpha_ix_i: \alpha_i>0\quad \forall\; i\right\}$$ is an open set.

Answer 1

Yes. You can take $\alpha=\operatorname{min}_i\alpha_i|x_i|$. Then take an open ball of radius $\alpha/2$.

If this contained a point, call $w=\sum_{i=1}^n\beta_ix_i$, not in the set, then $w$ has some $\beta_i\leq0$, which implies that $d(w,z)=\sum_{i=1}^n\left|\alpha_i-\beta_i\right||x_i|\geq(\alpha_i-\beta_i)|x_i|\geq\alpha_i|x_i|\geq\alpha>\alpha/2$, a contradiction.

Answer 2

Yes. The positive "octant" is always open.

One argument: the topologies generated by open rectangles and the one generated by open disks are the same (essence of proof: every open rectangle contains an open disk, and vice versa), and the positive octant is clearly a union of open rectangles (indeed, it's a single open rectangle).

(You might argue that you're not asking about the positive octant, but about the positive convex combination of these vectors. But there's an isomorphism taking those vectors to the standard basis vectors; that isomorphism is bicontinuous (being represented by multiplication by an invertible matrix, for instance), hence preserves open-ness.)

Answer 3

Yet another version of the same argument: The projections $\pi_i\colon\mathbb{R}^n\rightarrow \mathbb{R}$ given by $\pi_i(\sum_k{\alpha_kx_k})=\alpha_i$ are continuous, thus $\pi_i^{-1}((0,\infty))\subset \mathbb{R}^n$ is open for all $i=1,...,n$. The set under consideration is the interesection of these sets and thus open as well.