Convex sets and algebraic operations If $A$ is convex set and $\alpha,\beta>0$, show that $(\alpha+\beta)Α=\alpha Α+\beta Α$.
I tried to show that, but I am not sure if it was so simple. This is how I did it:
$$(\alpha+\beta)Α := \left\{z:z=(\alpha+\beta)x,\,x\in A\right\}$$
$\alpha Α := \left\{z:z=\alpha x, x\in A\right\}$
$\beta Α :=\left\{z:z=\beta x,\, x\in A\right\}$
Let it be $z\in (\alpha+\beta)Α$, $x\in A$. 
$$z=(\alpha+\beta)x=\alpha x+\beta x,$$ 
$\alpha x \in \alphaΑ$, $\beta x \in\betaΑ$.

The inclusion $(\alpha+\beta)A\subseteq \alpha A+\beta A$ is proved by your arguments, but the other part is yet missing: for that, let $x$ and $y$ be elements of $A$, and we need to prove that $\alpha x+\beta y\in (\alpha+\beta)A$.
For that, observe that, by convexity, we have
$$\frac{\alpha}{\alpha+\beta}x+\frac{\beta}{\alpha+\beta}y\in A\,.$$
