Show that there exists a sequence $ (a_n)\subset T$ for which $|a_n-a| \lneq 1/3^n $ for all n

Question

Let $T \subset R$, Let $ a= sup(T) $, and $ a\notin T $. Then Show that there exists a sequence $ (a_n)\subset T$ for which $|a_n-a| \lneq 1/3^n $ for all n.

I know it is not too hard to find a sequence which converges to a, but I get to know how to find such $a_n $sequence which satisfy $|a_n-a| \lneq 1/3^n $ for all n.

Answer 1

Let $(f(n))_{n\in \mathbb N}$ be any sequence of positive numbers converging to $0. $ Let $a=\sup T.$

From the def'n of $\sup,$ no $b<a$ is an upper bound for $T.$ So for each $n\in \mathbb N$ there exists $a_n\in T$ with $a_n>a-f(n).$

And also $a_n\leq a$ (because $a_n\in T$ and $a$ is an upper bound for $T.$)

So $\;a-f(n)<a_n\leq a,$ which implies $\;0\leq |a-a_n|=a-a_n<f(n).$

In particular we may take $f(n)=3^{-n}$ for each $n\in \mathbb N.$