Let $K \subset \mathbb{R}^n$ be a cone.
A cone is defined as: $(\forall x\in K) \wedge (\forall \lambda>0) \Longrightarrow \lambda x\in K$.
Prove that: $K$ is convex $\Longleftrightarrow K$ is closed under addition.
ATTEMPT
Part 1 $(\Longrightarrow)$
Let $x,y\in K$.
$K$ is convex, therefore: $\forall \lambda \in [0,1]$
$$\lambda x+(1-\lambda)y \in K \tag{1}$$
$K$ is a cone, therefore: $\forall \lambda,\lambda'\in (0,1]$ $$\lambda x,\lambda'y \in K \tag{2}$$
Let $\lambda'=1-\lambda$, then by $(1)$ we have: $\forall \lambda,\lambda'\in (0,1]$ $$\lambda x + \lambda 'y \in K \tag{3}$$
Then by $(2)$ and $(3)$ we have:
$$(\forall u=\lambda x \in K)\wedge (\forall v=\lambda'y \in K)\Longrightarrow u+v\in K$$
As desired.
Part 2 $(\Longleftarrow)$
Let $x,y\in K$.
$K$ is a cone, therefore: $\forall \lambda,\lambda'>0$ $$\lambda x,\lambda'y \in K \tag{4}$$
$K$ is closed under addition, therefore: $$\lambda x + \lambda'y \in K \tag{5}$$
Let $\lambda'=1-\lambda$, then $(5)$ is true for $\lambda\in (0,1)$ as in $(5)$ and is true for $\lambda \in \{0,1\}$ as in $(4)$.
Therefore: $\forall \lambda \in [0,1]$ $$\lambda x + (1-\lambda)y \in K \tag{6}$$
As desired.
I am still new to the concept of convexity and I do not trust my proof attempt.
Could anyone identify mistakes or provide tips?
Thank you.
