How to find the convex hull of a given set? 
*

*$A=\{(0,0),(0,1),(1,0)\}$

*$B=\mathbb{Q}^2$

*$C=\{(x,\sqrt{x})\in \mathbb{R}^2:x\ge0\}$


I have to find Conv(A), Conv(B) and Conv(C).
My attempt
Conv(A) is the boundary (correction: obviously it isn't just the boundary) of the triangle that has vertices in $(0,0),(0,1),(1,0)$ 
Conv(B) is $\mathbb{R}^2$ (we use the fact, that $\mathbb{Q}^2$ is dense in $\mathbb{R}^2$.
And I'm not sure about the last one, but my guess is:
Conv(C) $= \{(x,y)\in\mathbb{R}^2:x\ge0, \:\: 0\le y\le\sqrt{x}\}$
Is that correct? If so, how can I formally show it?
 A: In the following, we shall work with the following definition of the convex hull of a set $B$ in a vector space $V$:
Def:
Let $V$ be a vector space, and let $B \subseteq V$. $P \subseteq V$ is called the convex hull of $B$ iff $P$ is a convex set such that


*

*$B \subseteq P$

*for all convex sets $Q \subseteq V$ such that $B \subseteq Q$ we have $P \subseteq Q$


OK, so now let's start with the formal proofs.
First convex hull
We prove that the set
$$
A_1 := \{(x, y) \in \mathbb R^2 \big| x, y \ge 0 \wedge |x| + |y| \le 1\}
$$
is the convex hull of the set $\{(0, 0), (0, 1), (1, 0)\}$. So first, we note that $\{(0, 0), (0, 1), (1, 0)\} \subseteq A_1$. Second, we note that $A_1$ is convex, since for any $(x, y), (z, w) \in A_1$ and $\lambda \in [0, 1]$, we have
$$
\lambda x + (1 - \lambda) z, \lambda y + (1 - \lambda) w \ge 0
$$
and
$$
|\lambda x + (1 - \lambda) z| + |\lambda y + (1 - \lambda) w| \le \lambda(|x| + |y|) + (1 - \lambda) (|z| + |w|) \le 1
$$
Third, let $Q$ be any convex set containing $\{(0, 0), (0, 1), (1, 0)\}$. We show that every element of $A_1$ is in $Q$. Let thus $(x, y) \in A_1$ be arbitrary. Then we have $|x| + |y| \le 1$ and $|x| = x$, $|y| = y$. Therefore, the two elements $(0, |x| + |y|)$ and $(|x| + |y|, 0)$ are also contained in $Q$ as a convex combination of $(0, 1)$ and $(0,0)$ or $(0, 1)$ and $(0,0)$ respectively. Now either $x = y = 0$ and thus from $\{(0, 0), (0, 1), (1, 0)\} \subset Q$ automatically follows $(x, y) \in Q$ or we choose $\lambda = \frac{x}{x + y} \in [0, 1]$ and find
$$
\begin{align}
\lambda(|x| + |y|) & = x \\
(1 - \lambda)(|x| + |y|) & = y \\
\end{align}
$$
, which is why $(x, y) \in Q$, as $Q$ was supposed to be convex. $\Box$
Second convex hull
Because $\mathbb Q \subset \mathbb R$, it follows that $\mathbb Q^2 \subset \mathbb R^2$. Further, since $\mathbb R^2$ is a vector space, it is convex. Now let $Q$ be an arbitrary convex superset of $\mathbb Q^2$. We show that $\mathbb R^2 \subseteq Q$. Let $(x, y) \in \mathbb R^2$. Of course, the four elements $(\lfloor x \rfloor, \lfloor y \rfloor)$, $(\lfloor x \rfloor + 2, \lfloor y \rfloor)$ and $(\lfloor x \rfloor, \lfloor y \rfloor + 2)$ are contained in $\mathbb Q^2$ (and thus in $Q$). Since $0 \le x - \lfloor x \rfloor + y - \lfloor y \rfloor < 2$, if we choose $\lambda_1 := \frac{x - \lfloor x \rfloor + y - \lfloor y \rfloor}{2}$, then $\lambda_1 \in [0, 1]$. Further,
$$
\begin{align}
\lambda_1 (\lfloor x \rfloor + 2, \lfloor y \rfloor) + (1 - \lambda_1) (\lfloor x \rfloor, \lfloor y \rfloor) & = (x + y - \lfloor y \rfloor, \lfloor y \rfloor) \\
\lambda_1 (\lfloor x \rfloor, \lfloor y \rfloor + 2) + (1 - \lambda_1) (\lfloor x \rfloor, \lfloor y \rfloor) & = (\lfloor x \rfloor, y + x - \lfloor x \rfloor) \\
\end{align}
$$
Now either both $\lfloor x \rfloor = x$ and $\lfloor y \rfloor = y$ and thus $(x, y) \in \mathbb Q^2 \subseteq Q$, or we now choose $\lambda_2 := \frac{x - \lfloor x \rfloor}{x - \lfloor x \rfloor + y - \lfloor y \rfloor}$. Then
$$
\lambda_2 (x + y - \lfloor y \rfloor, \lfloor y \rfloor) + (1 - \lambda_2) (\lfloor x \rfloor, y + x - \lfloor x \rfloor) = (x, y)
$$
As $Q$ was supposed to be convex, all these elements, including the last one, are in $Q$. This shows that $\mathbb R^2 \subseteq Q$ and thus $\mathbb R^2$ is the convex hull of $\mathbb Q^2$. $\Box$
Third convex hull
We show that
$$
A_2 := \{(x, y) \in \mathbb R^2 \big| x, y > 0 \wedge y \le \sqrt{x} \} \cup \{(0, 0)\}
$$
is the convex hull of the set you denoted by $C$. It is easy to see that $C \subset A_2$. Further, $A_2$ is convex: Let $(x, y), (z, w) \in A_2$, $\lambda \in [0, 1]$. For $(x, y) = (z, w)$, clearly $\lambda (x, y) + (1 - \lambda)(z, w) = (x, y) \in A_2$. Otherwise $\lambda x + (1 - \lambda) z > 0$ and $0 < \lambda y + (1 - \lambda) w \le \lambda \sqrt{y} + (1 - \lambda) \sqrt{w} \le \sqrt{\lambda y + (1 - \lambda) w}$, where in the last $\le$ we used that $x \mapsto \sqrt{x}$ is a concave function. Thus, $\lambda (x, y) + (1 - \lambda)(z, w) \in A_2$.
Third, let now $Q$ be any convex set such that $C \subset Q$. We show that $A_2 \subseteq Q$. Let $(x, y) \in A_2$ be arbitrary. The first possibility is $(x, y) = (0, 0)$. In this case, $(x, y) \in C \subset Q$ and thus $(x, y) \in Q$. The second possibility is $(x, y) \neq (0, 0)$. In this case, the point $\left( \frac{x^2}{y^2}, \frac{x}{y} \right)$ is well-defined and contained in $C$. Further, of course, $(0, 0) \in C$. Since $Q$ contains $C$, of course also $(0, 0), \left( \frac{x^2}{y^2}, \frac{x}{y} \right) \in Q$. We choose now $\lambda := \frac{y^2}{x}$, and observe that since $(x, y) \in A_2$, $\lambda \in [0, 1]$. Then $\lambda \left( \frac{x^2}{y^2}, \frac{x}{y} \right) + (1 - \lambda) (0, 0) = (x, y) \in Q$ due to convexity of $Q$. Thus $A_2 \subseteq Q$ and $A_2$ is the convex hull of $C$. $\Box$
General recipe


*

*Try to visualize the convex hull in your inner eye and guess it.

*Show that your guess contains the set of which you wanted to find the convex hull.

*Prove that your guess is convex.

*Prove that any convex set containing the set will include your guess (here you will find any imprecisions your guess contained).

