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If $b_n \rightarrow b$ as $n\rightarrow \infty$ and $a \leq b_n \forall n$, show $a\leq b$

Proof: Let $\epsilon > 0$, since $b_n\rightarrow b$ as $n\rightarrow \infty$, there exists an $N_0\in \mathbb{N}$ such that $$N\geq N_0 \Rightarrow |b_n - b| < \epsilon$$

I want to state something like if $a$ is contained in the sequences $b_n$ then $a$ must be the minimal element of that sequence. But I am not sure that is an appropriate approach, any suggestions are greatly appreciated.

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1 Answer

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Assume toward contradiction that $a>b$. Notice that $a-b>0$, so let $\varepsilon=a-b$.

Then there exists some $N_0$ such that $n\geq N_0$ implies $|b_n-b|<\varepsilon=a-b$. Obviously, then: $$b_n-b<a-b$$ which is directly in contradiction with the assumption that $a\leq b_n$ for all $n$.

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