Let $P$ be a polyhedron in $\mathbb{R}^{n}$ defined by $$\langle a_{i},x\rangle \leq b_{i}, i \in [k], \textrm{ or equivalently, } P = \{x \in \mathbb{R}^{n}: Ax\leq b\}$$ and let $L:\mathbb{R}^{n} \to \mathbb{R}^{m}$ be a linear transformation. Then, $$L(P_{\infty}) = [L(P)]_{\infty},$$ where $P_{\infty} = \{y \in \mathbb{R}^{n}:Ay \leq 0 \}.$
This is the corollary 2.3.17 in the springer's book, "Multiobjective Linear Programming." He uses below argument.
To see $L(P_{\infty}) \subseteq [L(P)]_{\infty}$, let $u \in P_{\infty}.$ Then, for any $t >0, x\in P$, $x+tu \in P$. Thus, $L(x)+tL(u) \in L(P)$ for all $t \geq 0$. Hence, $L(u) \in [L(P)]_{\infty}$.
To see converse inclusion, let $v \in [L(P)]_{\infty}$ with $v\neq0$. Then, for any $x \in P, t>0$, $$L(x) + tv \in [L(P)]_{\infty}.$$ Now choose $\{x_{i}\}_{i \in \mathbb{N}} \subseteq P$ such that $L(x_{i}) = L(x) + iv$ for each $i$. Such $x_{i}$ exists in $P$ by preceding sentence. So $$v = L\left(\frac{x_{i}-x}{i}\right), \forall i \in \mathbb{N}.$$
Now, he just assumes that
1) "Without loss of generality, we can have we may assume that the vectors $\frac{x_{i}-x}{i}$ converge to some nonzero vector $u$ as $i$ tends to $\infty$."
2)Then $$\langle a_{j},u\rangle = \lim_{i \to \infty} \left(\langle a_{j},\frac{x_{i}}{i}\rangle - \langle a_{j},\frac{x}{i}\rangle \right) \leq 0 \textrm{ for any } j \in [k]$$
Could you help me understand how we may assume 1) and derive 2) from the 1)?
EDIT: My question is that 1) cannot be justified since $P$ may not polytope, i.e., not compact in $\mathbb{R}^{n}$. Is there any result stating that a sequence in the convex region always has the convergent subsequence? I think this statement may not be true.
And still, even if I assume 1), I don't understand why $2)$ holds.