I want to show that if $$f: \Bbb R^n \to \Bbb R$$ is a convex function, then the set $$C=\{\ x\in \Bbb R^n : f(x) \le c\ \}$$ is a convex set.
My intuition tells me to do the following: Let $x_1, x_2 \in C$, then from here we know that $f(x_1)<C$ and $f(x_2)<C$. Now I want to kinda dorm the definition of the convex function by defining an $x=\lambda x_1 + (1 - \lambda)x_2$ for some $\lambda\in[0,1]$. From here I think I can finish it.
My question stems from the first part of my "proof". Can I just start right off the bat like that and pull elements out of the set? Do I have to show my "formulation" of the set or of the function?
Yes, you can pull $x_1,x_2\in C$ out without saying anything else. And yes, as $f$ is convex, you know that it satisfies the inequality convex functions satisfy.
This is the strength of abstract proofs - we specifically don't want to specify specific functions or sets. By not specifying an explicit function or set, this now holds for all convex functions $f:\mathbb R^n\to\mathbb R$ and all corresponding sets $C = \{x\in\mathbb R^n\mid f(x)\leq c\}$.