Let $C \subset R^n$ be a closed, convex cone. Let $T: R^n \to R^m$ be a linear transformation.
Is $T(C) \subseteq R^m$ closed?
I'm very positive that the answer is NO. But I couldn't come up to a counter example so far. I also realized that any counter example (if exists) has to be in Dimension $n \geq 3$, otherwise $C$ becomes a polyhedral and so is $T(C)$.
On
The question was already answered by @J_P. Here is another way of thinking: There are two closed convex cone, $C_1$ and $C_2$ in $R^3$ whose their sum i.e., $C_1 + C_2$ is not closed see "https://math.stackexchange.com/a/3273305/219176"
Clearly $C_1 \times C_2$ is closed convex cone. Now take the linear transformation $T(x,y)= x+y$ then we have $T(C_1 \times C_2) = C_1 + C_2$.
Big thanks to @Shalop who had the idea of using plain old cones as a counterexample, I just cleaned up the algebra.
So, by rotating the cone defined by $x^2+y^2=z^2;z\geq0$ by $45$ degrees toward the positive $x$-axis, we get the cone $y^2=2xz$ (this is just an exercise in $2D$ rotations). The projection of this cone to the $xy$ plane is the set $$S=\{(x,y):x>0\}\cup\{(0,0)\}$$ This is simple to see. If $x=0$, then $y^2=0$ and so only the vertical ray $(0,0,z);z\geq0$ get projected onto the $y$-axis onto the point $(0,0)$. If $x>0$, we get that the point $(x,y,y^2/2x)$ is on the cone and gets projected into $(x,y)$. This covers all possibilities, so the projection of the cone is $S$ but $S$ is not closed.