Is $C_1 + C_2$ closed?

Question

Two questions jump into my mind when I was working with cone, I feel they are very related to each other. one I asked here "Is $T(C) \subseteq R^m$ closed?"

Second one is: If $C_1$ and $C_2$ are two closed, convex cone in $R^n$ then is $$C_1 + C_2 = \{c_1 + c_2 \; | \; c_1 \in C_1 , ~ c_2 \in C_2\} $$

a closed set?

The answer for my first question is No, I feel the answer of my second question is No as well but I dont have an explicit example to confirm that.

Answer 1

OK, a counter-example.

Let $C_1$ be the closed convex cone $x^2+y^2\leq z^2$, $z\geq 0$ in $\mathbb{R}^3$, and $C_2$ be the half-line $t(1,0,-1)$, $t\geq 0$.

Then $C_1+C_2=\operatorname{conv}(C_1\cup C_2)$ does not contain $(0,1,0)$, but it contains $$ \overbrace{(-t,1+t^{-1},t + \frac{t^{-1} + 2t^{-2} + t^{-3}}2)}^{\in C_1}+\overbrace{(t,0,-t)}^{\in C_2}=(0,1+t^{-1},\frac{t^{-1} + 2t^{-2} + t^{-3}}2) $$ for all $t\gg 1$.