Is linear combination of a finite set is $\mathbb R^2$ when some condition is given? 
Let $A \subseteq \mathbb R^2$ be a finite set such that the set $(-\delta,\delta) \times (-\delta,\delta)$ is generated by  $A$ (in the sense of convex combinations). Then is it true that 'linear combination of elements of A is $\mathbb R^2$' ?
Here $\delta >o$ is some real number.

Actually in a lemma it has been used so I think it should be true but how?
Please someone help..
Thank you..
 A: Well, suppose $(x_1, y_1), \ldots, (x_n, y_n) \in \mathbb{R}^2$ with the property that
$$\operatorname{conv}((x_1, y_1), \ldots, (x_n, y_n)) \supseteq (-\delta, \delta)^2$$
Suppose $(x, y) \in \mathbb{R}^2$. If $(x, y) = (0, 0)$, then we are done, as $(0, 0) \in (-\delta, \delta)^2$. Otherwise, let $\alpha = \frac{\max(|x|, |y|)}{\delta}$. Then $(x, y)/\alpha \in (-\delta, \delta)^2$ (you should verify this). So, $(x, y)/\alpha$ can be expressed as a convex combination of $(x_1, y_1), \ldots, (x_n, y_n)$. That is, there exist $\lambda_1, \ldots, \lambda_n \in [0, \infty)$ such that $\sum_{i=1}^n \lambda_i = 1$ and
\begin{align*}
& (x, y)/\alpha = \lambda_1 (x_1, y_1) + \ldots + \lambda_n (x_n, y_n) \\
\implies& (x, y) = \alpha\lambda_1 (x_1, y_1) + \ldots + \alpha\lambda_n (x_n, y_n)
\end{align*}
which is in the span of the above vectors.
There's a more general version too: if the convex hull of a set of vectors are a neighbourhood of any point (that is, it contains an open ball of some radius), then they must span. See if you can prove it!
