A convex polyhedron is a set $S$ that satisfies $S = \{x \in \mathbb{R}^n :\ Ax\preceq b,\ Cx = d\}$ for some positive integers $m, n, p \in \{1,2,\dots\}$, some matrices $A \in \mathbb{R}^{m\times n}$, $C \in \mathbb{R}^{p\times n}$, and some vectors $b \in \mathbb{R}^m$ and $d \in \mathbb{R}^p$. (We write $v\preceq w$ for $v_1\leq w_1,\ v_2\leq w_2,\ \dots,\ v_q\leq w_q$ whenever $v, w \in \mathbb{R}^q$ are vectors, for some $q \in \{1,2,\dots\}$.)
Consider the following set. $S := \big\{x \in \mathbb{R}^3 :\ \forall t \in [1,\infty)\bullet x_1 + e^{-t}x_2 + e^{-2t}x_3 \leq 1.1\big\}$. Is $S$ a convex polyhedron?
Following is my attempt at solving this problem.
I represented $S$ as a union $S = S_{000}\cup S_{001}\cup S_{010}\cup\cdots \cup S_{111}$, where for every $\xi \in \{000,001,010,\dots,111\}$, $S_\xi := S\cap T_\xi$, with $$ \begin{align*} T_{000} &:= \{x \in \mathbb{R}^3 :\ x_1 \leq 0, x_2 \leq 0, x_3 \leq 0\}\\ T_{001} &:= \{x \in \mathbb{R}^3 :\ x_1 \leq 0, x_2 \leq 0, x_3 \geq 0\}\\ T_{010} &:= \{x \in \mathbb{R}^3 :\ x_1 \leq 0, x_2 \geq 0, x_3 \leq 0\}\\ &\vdots\\ T_{111} &:= \{x \in \mathbb{R}^3 :\ x_1 \geq 0, x_2 \geq 0, x_3 \geq 0\} \end{align*} $$ Note that every $T_\xi$ is a convex polyhedron, and that a finite intersection of convex polyhedra is a convex polyhedron.
I noticed that $S_{111}$ and $S_{000}$ were convex polyhedra: $$ \begin{align*} S_{111} &= \{x \in \mathbb{R}^3 :\ x_1 + e^{-1}x_2 + e^{-2}x_3 \leq 1.1\} \cap T_{111},\\ S_{000} &= T_{000}, \end{align*} $$ and wondered whether all the $S_\xi$'s were convex polyhedra, but was unable either to prove or to disprove this conjecture. However, even if I were able to prove it, it wouldn't in itself solve the question of whether $S$ was a convex polyhedron.