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> Let $\{a\_n\},\{b\_n\}$ be given such that $a\_n\leq b\_n$, then if $\{a\_n\}$ converges to $a$ and $\{b\_n\}$ converges to $b$, show that this implies $a\leq b$.
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Proof:

Let $\epsilon$. Since $\{a\_n\}$ converges to $a$ and $\{b\_n\}$ converges to $b$, then we have 
$$\vert a\_n-a\vert<\epsilon/2\,\,\,\,\text{for all $n\geq N$}\tag1$$ $$\vert b-b\_n\vert<\epsilon/2\,\,\,\,\text{for all $n\geq M$}\tag2$$ 

Add $(1)$ and $(2)$, we have $\vert a\_n-a\vert+\vert b-b\_n\vert<\epsilon$

Apply triangle inequality, we have $\vert(a\_n-a)+(b-b\_n)\vert<\epsilon$ and this is equivalent to $\vert(a\_n-b\_n)+(b-a)\vert<\epsilon$. Because $a\_n\leq b\_n$ and let $k\geq\max\{M,N\}$, then we get $$\vert a\_k-b\_k\vert-\vert b-a\vert\leq\vert(a\_k-b\_k)+(b-a)\vert\leq\vert a\_k-b\_k\vert+\vert b-a\vert$$

We know that $\vert a\_k-b\_k\vert+\vert b-a\vert\geq 0$, so $-\vert a\_k-b\_k\vert\leq\vert b-a\vert$ and this gives us $0\leq\vert b-a\vert$ and implies $a\leq b$.

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I have a feeling that I missed something that I need to show. Can anyone check my proof to see what I missed? Thanks.