I have a set of all bounded sequences of real numbers with the metric $d$ defined as:
d(a, b) = sup$_n|a_n - b_n|$.
I want to prove this is a complete metric space, i.e. every Cauchy sequence is convergent.
My attempt at Solution:
Consider a Cauchy sequence $a_n$. For every $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that for all $m, n > N$, we have sup$_{n, m} |a_n - a_m| < \epsilon$. $\implies $ $|a_n - a_m| < \epsilon$ for all $n, m > N$.
Every bounded sequence in reals has a convergent subsequence - call this subsequence $a_{n_k}$. Hence, for every $\epsilon > 0$, there exists an $M \in \mathbb{N}$ such that for all $k > M$, we have $|a_{n_k} - L| < \epsilon$ $\implies $ sup$_k |a_{n_k} - L| < \epsilon$.
Let $N^* = \max \{N, n_M \}$ so that for all $n, m > N^*$ and $n_k > N^*$ for some $k > M^*$, we have:
sup$_k|a_{n_k} - L| < \epsilon$ and
sup$_{n, m}|a_n - a_m | < \epsilon$.
Then,
$|a_n - L | = |a_n - a_{n_k} + a_{n_k} - L| \leq |a_n - a_{n_k}| + |a_{n_k} - L| $.
Now if $k > M^*$ such that $n_k > N^* \geq N$ and also $n > N^*$, we have
$|a_n - L| \leq 2 \epsilon$
Hence, sup$_n |a_n - L| \leq 2 \epsilon$.