# Infimum in a Sequence

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.

Hint: for large enough $$n$$ we have $$a_n>3$$. Consider the finitely many elements of the sequence before that point.
• 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 $\inf{a_n}=a_{n_1} > 2$ – talbi Dec 3 '20 at 17:37
• 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. – Olivier Moschetta Dec 3 '20 at 22:30
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 $$n$$, $$4 so there exist only other $$N-1$$ possible terms that are under $$4$$.
This means that $$\inf a_n = \inf_{n 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 $$n_0 such that $$a_{n_0}=2$$.