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Let $a_n$ be a sequence such that $$ \inf{a_n} = 2 $$ and $$ \lim{a_n} = 5 $$

Does exist $n_0$ such that $a_{n_0} = 2$?

I'm not sure how to approach this question, I believe that $2$ is in $a_n$, since there only exists one partial limit, and so if $2$ wasn't in $a_n$, then $\inf{a_n} > 2$ (since $a_n$ must be increasing?).

I'm not sure how I could justify myself, or if my argument is correct.

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Hint: for large enough $n$ we have $a_n>3$. Consider the finitely many elements of the sequence before that point.

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  • $\begingroup$ Got it. So then, I can say that if $n_0$ doesn't exist, then $\inf{a_n} > 2$, right? Since $n_0$ must be finite, then if it doesn't exist, then exists $n_1$ in the same range such that $2 < a_{n_1} \leq a_n$, and so $\inf{a_n}=a_{n_1} > 2$ $\endgroup$ – talbi Dec 3 '20 at 17:37
  • $\begingroup$ Yes if there is no $n_0$ such that $a_{n_0}=2$ then the inf over the finitely many first elements has to be $>2$ (since the inf is always attained on a finite set), and the inf over the rest is at least 3. $\endgroup$ – Olivier Moschetta Dec 3 '20 at 22:30
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By definition $$\forall \epsilon>0 \mbox{, } \exists N \mbox{ s.t. } \forall n\ge N \mbox{, }|a_n-5|<\epsilon$$ so given $\epsilon=1$, $$\exists N \mbox{ s.t. } \forall n\ge N \mbox{, }|a_n-5|<1$$ so for all of these $n$, $4<a_{n}$ so there exist only other $N-1$ possible terms that are under $4$.

This means that $\inf a_n = \inf_{n<N}a_n=\min_{n<n_0}a_n=2 $ becouse of the finiteness of the set on which we are taking the $\inf$, so there exists by definition an element $a_{n_0}$ with $n_0<N$ such that $a_{n_0}=2$.

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