My working to this question is implies that $a$ is indeed the inf, but I am not really sure whether $\{a_n\} \downarrow a$ implies that $a_n \ge a \quad \forall n$ since I am really not used to the down arrow notation.
$a_n \downarrow a$, is $a$ the $\inf$ of the sequence?
0
$\begingroup$
analysis
-
0The meaning of this notation is that $a_n\leq a_{n-1}$. What happens if for some $n$, a_n? – 2012-11-03
1 Answers
1
If $a_n \downarrow a$, then $a_n\geq a_{n+1}$, and also $a_n\geq a$ for all $n$ (otherwise an immediate contradiction). Hence $\inf_n a_n \geq a$. Let $\epsilon>0$, then there is some $n$ such that $|a_n -a| < \epsilon$, in particular, $a_n< a + \epsilon$, hence $\inf_n a_n \leq a + \epsilon$. It follows that $\inf_n a_n = a$.
To see why $a_n\geq a$, suppose $a_{n'} < a$ for some $n'$, then for $n\geq n'$, $|a-a_n| > a-a_{n'} > 0$ which contradicts $a = \lim_n a_n$. Hence $a_n \geq a$ for all $n$.