d(($a_n),(b_n))$ in $l^{\infty}$

Question

Working with $l^{\infty}$ as a space of all bounded sequences $(a_n)$ $\epsilon$ $\mathbb{R}$ defined such that $||(a_n)||_{\infty}$ = sup{$|a_n|$ : $n$ $\epsilon$ $\mathbb{N}$ }.

I'm trying to show that d(($a_n),(b_n$)) is a finite number.

Can I say that for all sequences $(a_n)$ there must exist some $a_i, b_j$ such that d($(a_i),(b_j)) < \epsilon$ for some $\epsilon>0$ since the sequences are defined as bounded, thus as ||$(a_i-b_j)|| < \epsilon$ then there does exist a finite number for some $a_i,b_j$ in $l^{\infty}$?

Answer 1

The space $\ell^\infty = \{(x_n)_n \in \mathbb{R}^\mathbb{N}: \exists M \in \mathbb{R} : \forall n: |x_n| \le M\}$ of all bounded real sequences, is a linear space: the all zero-sequence is certainly bounded, the same $M$ that works for $x$ also works for $-x$, and $|x + y| \le |x| + |y|$ so if both terms on the right are bounded, so is the sum.

By my definition it follows that $||x||_\infty := \sup \{|x_n|: n \in \mathbb{N}\}$ is well-defined and finite (we have a subset of $\mathbb{R}$ that is bounded above by some $M$, so its sup exists and is $\le M$ as well. (this uses the order completeness of the reals that a lot of analysis relies upon)

Now, if the norm is well-defined, so is its derived metric: $(a_n)_n, (b_n)_n \in \ell^\infty$ implies $((a_n - b_n)_n \in \ell^\infty$, so $d((a_n)_n, (b_n)_n) = ||a_n - b_n||_\infty < \infty$