Defining a set $C \subset \mathbb{R}^n$ as a cone if for ever $x \in C$ and $\alpha \geq 0$
we have $\alpha x \in C$. ie they are closed under scalar multiplication.
How can I show that the intersection and unions of a cone is also a cone?
I am new to mathematical proofs so am not sure if I have the correct approach, but this is what I tried:
Let $C_1$ and $C_2$ be two cones.
The domain of $C_1 \cap C_2 $ = $C_1 \cap C_2$
Then $\forall $ $x\in C_1$, $y\in C_2$ and $\forall \alpha ,\beta \geq 0$
$ \alpha x + \beta y \in C_1 \cap C_2$
Hence the intersection of cones is a cone. Now I only showed this for the case of two intersections, but could I use a similar method to prove the intersection of n cones is a cone? Have I sufficiently proved this statement?
For the union I did dom$C_1 \cup C_2$ = $C_1 \cap C_2$ and did the same thing. IS this correct?
Best Answer
First, show that the intersection $C_1\cap C_2$ is a cone: take $\alpha\ge0$, $x\in C_1\cap C_2$. Then $x\in C_1$ and $x\in C_2$, and since both are cones $\alpha x\in C_1$ and $\alpha x\in C_2$.
To prove that the intersection of $n$ cones is a cone, use induction with the above argument in the induction step.
The proof for union of cones is a cone, use a similar argument.