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Let $(a_n)$ and $(b_n)$ be sequences of real numbers that converges to real numbers $a$ and $b$, respectively. If $a_n\leq b_n$ for all $n\in\mathbb{N}$, then $a\leq b$.

Could I prove this by contradiction and assume $a>b$?

I am unsure how to approach this.

Ryan
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1 Answers1

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You assume $a>b$. Now you may proceed as follows:

  • Since the distance from $a$ to $b$ is $a-b$, you choose an $\epsilon < \frac{a-b}{2}$.

Since $\lim_{n\to\infty}a_n = a$ and $\lim_{n\to\infty}b_n = b$ there exists $N \in \mathbb{N}$ large enough such that

  • $a_N> a-\epsilon$ and $b_N < b+\epsilon$

Putting this together you get $$a_N > a -\epsilon > a-\frac{a-b}{2} = \frac{a+b}{2}$$ $$b_N < b +\epsilon < b+\frac{a-b}{2} = \frac{a+b}{2}$$ All together $$a_N > \frac{a+b}{2} > b_N \mbox{ which contradicts to } a_n \leq b_n \mbox{ for all } n\in\mathbb{N}$$

trancelocation
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