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Show that every open $n$-ball contains $n$ vectors that form a basis for $\mathbb{R}^n$.

Any hints on how to solve this?

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I don't know which level of formalism is expected, and the context - but is this clear to you? If not, note that you can map each n-ball bijectively to the unit sphere around the origin (maybe you have seen this; if not, it is not hard to see and show). Once there, look at the 3 dimensional case: the three unit vectors spanning the Euclidian 3-dimensional sphere are elements of the ball now. They span $\mathbb{R^3}$. The general case is just the same. – gnometorule Jan 28 '13 at 6:58

$\newcommand{\eps}{\varepsilon}$Hint: Let $B(x,\eps)$ be an open ball and $e_1,\dots,e_n$ be the standard unit vectors. What happens if you look at something like $x+e_1\eps/2$, careful though that you don't hit zero.

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Hitting zero is one problem, but not the only problem. Fortunately, such a choice will work for all but finitely many choices of $\varepsilon$, as made clear in copper.hat's answer. – Jonas Meyer Jan 28 '13 at 19:33

If $B$ is an open ball, then $U=B+(-B)$ is an open neighborhood of $0$, which implies that $\mathbb R^n=\bigcup\limits_{k=1}^\infty k\cdot U$. This shows that $\mathbb R^n$ is in the span of $B$. Every spanning set contains a basis.

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Let $U \subset \mathbb{R}^n$ be open and $x \in U$. Then $B(x,\epsilon) \subset U$ for some $\epsilon>0$. Consider the function $\eta(r) = \det \begin{bmatrix}x+ r e_1& \cdots & x + r e_n\end{bmatrix} = \det (\begin{bmatrix}x& \cdots & x \end{bmatrix} +r I )$. $\eta$ is a degree $n$ polynomial in $r$, hence has at most $n$ zeroes. Consequently, we can find an $r \in (0,\epsilon)$ such that $\eta(r) \neq 0$. It follows from the properties of the determinant that the vectors $x+r e_k$ are linearly independent, hence form a basis for $\mathbb{R}^n$.

Here is an alternative approach based on measure: Let $\phi:\mathbb{R}^n \times \cdots \mathbb{R}^n \to \mathbb{R}$ defined by $\phi(x_1,...,x_n) = \det \begin{bmatrix}x_1& \cdots & x_n\end{bmatrix}$. The set $\phi^{-1} \{0\}$ has Lebesgue measure zero (see, eg, Probability of having zero determinant) and the set $U\times \cdots \times U \subset \mathbb{R}^n \times \cdots \mathbb{R}^n$ is open (and so has strictly positive measure), hence $\phi(x_1,...,x_n) \neq 0$ for almost all points $(x_1,...,x_n) \in U\times \cdots \times U $. It follows that almost any selection of $n$ points in the ball will be linearly independent, hence forming a basis for $\mathbb{R}^n$.

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It suffices to show that if $B$ is an open ball then $\text{Span}(B)=\mathbb R^n$. To do this let $a$ and $r >0$ center and radius of $B$ and $x \in \Bbb R^n$

If $x=a$ then $x \in \text{Span}(B)$ ( because $B \subset \text{Span}(B)$ )

If $x \neq a$ we have : $x=a+ \frac{2\|x-a\|}{r}\frac{r}{2\|x-a\|}(x-a) \in \text{Span}(B)$ since $\in B$ and $a \in B$ and $\frac{r}{2\|x-a\|}(x-a) = \underbrace{\frac{r}{2\|x-a\|}(x-a) + a}_{\in B} -a$

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