I'm attempting to prove the following theorem.
Theorem. Let $\langle a_n \rangle$ be some arbitrary, bounded sequence, and let $A_n$ denote the set $\{a_n, a_{n+1}, a_{n+2}, \ldots\}$. Suppose $\langle a_n \rangle$ converges to some $a$, and let $\langle G_n \rangle$ be a sequence where the $G_n^{th}$ term gives the greatest lower bound of the set $A_n$. Show that $\langle G_n \rangle$ converges to $a$ as well.
Note it can be shown that $\langle G_n \rangle$ is increasing and convergent to $lub(\langle G_n \rangle)$ (where $lub(A)$/$lub\{A\}$ denotes the least upper bound of $A$ (this follows from the fact that a bounded, increasing monotonic sequence is convergent to its least upper bound).
Attempt. Suppose that $\langle a_n \rangle$ converges to $a$. Then, for all real $\epsilon >0$, there exists $N\in\mathbb{N}$ such that, for all $n\geq N$:
$$-\epsilon < a_n – a < \epsilon$$
Further, it is clear that, for all $n\in\mathbb{N}$,
$$G_n \leq a_n$$
…for $\langle G_n \rangle$ is the greatest lower bound of the set $\{a_n, a_{n+1}, a_{n+2}, \ldots \}$.
Hence, for $n$ at least $N$:
$$G_n – a \leq a_n – a < \epsilon$$
Since $\langle G_n \rangle$ converges to $lub\{\langle G_n \rangle\}$, we know also that, for all $\epsilon>0$, there exists $M \in \mathbb{N}$ such that, for all $n\geq M$:
$$-\epsilon < G_n – lub\{\langle G_n \rangle \} < \epsilon$$
It is clear that, for all $n\in\mathbb{N}$:
$$G_n – lub\{\langle G_n \rangle \} \leq G_n -a$$
For otherwise:
$$G_n – lub\{\langle G_n \rangle \} > G_n -a \textrm{ } \Rightarrow \textrm{ } – lub\{\langle G_n \rangle \} > -a \textrm{ } \Rightarrow \textrm{ } lub\{\langle G_n \rangle \} < a$$
…here's where I'm stuck. I had in mind an argument for why the last line leads to a contradiction (or alternatively why $lub\{\langle G_g \rangle \}$ must be $a$), but I've forgotten it (if it was even valid).
Am I headed in the right direction? How may I proceed?
Best Answer
You have used the fact that $G_n$ is a lower bound of $A_n$. Since not necessarily all sequence of lower bounds of $A_n$ converge to $a$, you should somehow use the fact that $G_n$ is the 'greatest' lower bound of $A_n$.
For each $n\ge N$, since $G_n$ is the greatest lower bound of $A_n$, there is some $m\ge n$, s.t $G_n \le a_m\le G_n+\varepsilon$. Since $a-\varepsilon < a_n < a+\varepsilon$ for all $n\ge N$, we have $$a-2\varepsilon<a_n-\varepsilon\le G_n\le a_n<a+\varepsilon.$$ Since this holds for all $n\ge N$, by definition of limit we have shown that $G_n\to a$.