Convex set. Proof. Prove that if A is convex set and $\alpha, \beta ≥ 0$ then $(\alpha + \beta)A = \alpha A + \beta A$
What came first on my mind is that I have to show that $(\alpha + \beta)A\subset \alpha A + \beta A$ and $\alpha A + \beta A\subset (\alpha + \beta)A$. But I don't know how to use convexity of A to show this. 
 A: One direction doesn't require convexity.  Let $(\alpha+\beta)a\in (\alpha+\beta)A$.  Then $\alpha a\in \alpha A$ and $\beta a\in \beta A$, and $(\alpha+\beta)a=\alpha a+\beta a$, so $(\alpha+\beta)a\in \alpha A + \beta A$.
For the other direction, let $\alpha a_1 + \beta a_2\in \alpha A + \beta A$.  We need to find some $a_3\in A$ such that $\alpha a_1+\beta a_2=(\alpha+\beta)a_3$.  To find such an $a_3$, divide both sides by $(\alpha+\beta)$. (footnote*).  We set $a_3=\frac{\alpha}{\alpha+\beta}a_1+\frac{\beta}{\alpha+\beta}a_2$.  Because $A$ is convex and these fractions are in $(0,1)$ and add up to 1, $a_3\in A$ and we're done.
(footnote*) This proof frays at the edges if $\alpha=\beta=0$, so this special case must be handle separately; it's easy though since $(0+0)A=0A+0A=\{0\}$ for any $A$.
A: The "$\subset$" part is easy: $(\alpha + \beta)x=\alpha x + \beta x \in \alpha A +\beta A \ \forall x \in A$;  on the other hand $\alpha x + \beta y=(\alpha+ \beta)(\frac{\alpha x}{\alpha + \beta} + \frac {\beta y}{\alpha + \beta})\in(\alpha+\beta)A$  where the membership relation is due to the convexity of A.
A: For the open half-space (I suppose in an n-vector space over the real numbers), consider that up to an isomorphism they're $\{ (x_1,...,x_n): x_n >0\} $ so that each convex combination remains into that set.
A: Could it be possible to try the definition of convex sets? 
$A$ is a convex set and $\alpha,\beta\geq 0$, then $(\alpha+\beta)A=\alpha A+\beta A$. As $\alpha A$ and $\beta A$ in $A$, then according to the definition of convexity $\lambda\alpha A + (1-\lambda)\beta A$ equals $(\lambda\alpha + (1-\lambda)\beta)A$. This implies $A$ inverse $(\lambda\alpha + (1-\lambda)\beta)$ equals $\lambda\alpha + (1-\lambda)\beta$, and that is element $A$ so it exist.
