**Lemma** If there exists $N \in \mathbb{N}$ such that $a_{n} \leq b_{n}$ for all $n \geq N$, $a_{n} \to a$ and $b_{n} \to b$ then $a \leq b$ (i.e [[Convergence|limits]] preserve weak inequalities). **Proof**: BWOC suppose $a>b$. Choose $\epsilon=(a-b)/2$; then $\exists M$ such that $\forall n \geq M$, $|a_{n}-a|< \frac{a-b}{2} \text{ and } |b_{n}-b|< \frac{a-b}{2}$so then $a-a_{n} \leq |a_{n}-a|< \frac{a-b}{2} \implies a_{n} > a - \frac{a-b}{2 } = \frac{a+b}{2}$and $b_{n}-b \leq |b_{n}-b | < \frac{a-b}{2} \implies b_{n} < \frac{a+b}{2} $If we take some $n > \max(M,N)$ this contradicts the assumption that $a_{n} \leq b_{n}$ for all $n \geq N$.