A cone in $\mathbb{R}^n$ containing n linearly independent vectors has a non empty interior I need a help with proving, that if a cone $K \subseteq \mathbb{R}^n$ contains $n$ linearly independent vectors, then the interior of $K$ is non empty.
Lets say $b_1,\dots,b_n \in K$ are the independent vectors. From linear algebra we know that they form a basis of $\mathbb{R}^n$. By definition of "non-emptiness" I just need to find a point $p \in K$ and $\epsilon > 0 $ such that the ball $B = \{x \in \mathbb{R}^n: \Vert x-p \Vert_2 
 < \epsilon \}$ is contained in $K$. 
By intuition in "2D" I think it should work if I take $p = \sum_{i=1}^n b_i $. $p$ is a conic combination of vectors in $K$, so $p \in K$. Now I just need to take $y \in B$ and prove that the coordinates $a_1,\dots,a_n \in \mathbb{R}$ of $y = \sum_{i=1}^n a_i b_i$ in basis $b_1,\dots,b_n$ are non-negative. That would mean that $y \in K$ therefore $B \subseteq K$. 
I just cannot seem to find a way how to prove that $a_i \geq 0$. From the ball definition I just have that $\Vert \sum_{i=1}^n (1-a_i)b_i \Vert_2 < \epsilon$ but thats it.
 A: The map $S^{n-1}\to[0,\infty)$, $(u_1,\ldots, u_n)\mapsto\|u_1b_1+\cdots + u_nb_n\|$ is continuous and hence attains its minimum $r$, say. By linear independence, $r\ne0$. Then every $v\in\Bbb R^n$ with $\|v\|<r$ can be written as some $\frac{\|v\|}{r}(u_1b_1+\cdots + u_nb_n)$, i.e., all points in the open $r$-ball around $0$ have all their coordinates (with respect to basis $b_1,\ldots,b_n$) absolutely $<r$.
Note that the conic combination $v_0:=\frac1r(b_1+\cdots +b_n)$ is $\in K$. By what we just found, every point in the open $r$-ball around $v_0$  has all coordinates (with respect to basis $b_1,\ldots,b_n$) in the interval $(0,2r)$ and hence is also a conic combination of the $b_i$.
A: The result depends on the definition of cone we use, so I will assume that $K$ is a convex cone (otherwise, $K$ might reduce to a set of halflines, which has no interior if $n>1$).
Since the conic hull generated by $b_1,\dots,b_n$ is included in $K$, it is enough to show that this cone has a non-empty interior.
Take any convex combination
$$
   v = \sum_{i=1}^n\alpha_ib_i,
$$
where all coefficients $\alpha_i>0$, and let's show that $v$ is in the interior of the cone.
Assume it is not. Then there would exist a sequence $(v^{(k)})_k$ of vectors not in the cone such that
$$
   \lim_k v^{(k)} = v.
$$
Write every $v^{(k)}$ as a linear combination of $b_1,\dots,b_n$
$$
   v^{(k)} = \sum_{i=1}^n\alpha_{i,k}b_i.
$$
Since no $v^{(k)}$ is in the cone, for every $k$ there must exist at least one index $i_k$ such that $\alpha_{i_k,k}<0$. Using the fact that there are only $n$ indexes $i$ and an infinite number of indexes $k$, there must be an index $i_0$ such that $\alpha_{i_0,k}<0$ for an infinite number of indexes $k$. Take the subsequence $(k_j)_j$ of such indexes.
The function $\phi_i$ that selects the $i$th coordinate of a vector in a base is a linear form, and consequently continuous. Applying $\phi_{i_0}$ to $v$ we get
$$
    0 < \alpha_{i_0} = \phi_{i_0}(v) = \phi_{i_0}(\lim_jv^{(k_j)})= \lim_j\phi_{i_0}(v^{(k_j)}) = \lim_j \alpha_{i_0,k_j} \le 0,
$$
which is a contradiction. Then $v$ is in the interior of the cone, as desired.
A: Since $b_1,\dots,b_n$ is a basis of $\mathbb R^n$, the linear map $\varphi\colon\mathbb R^n\to\mathbb R^n$ defined by $\varphi(b_i)=e_i$, where $e_1,\dots,e_n$ are the standard basis vectors, is a linear bijection and hence in particular a homeomorphism. Hence, $K$ has non-empty interior if and only if $\varphi(K)$ has non-empty interior. But $\varphi(K)$ contains $(\mathbb R_{\ge 0})^n$ which in particular contains the open set $(\mathbb R_{>0})^n$. Alternatively, just find a ball around $(1,1,\dots,1)$ in $(\mathbb R_{\ge 0})^n$.
