1
$\begingroup$

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 $l^{\infty}$?

$\endgroup$
  • 1
    $\begingroup$ If you are taking $d(a_n, b_n)=\lVert a_n-b_n \rVert_\infty$, then you already have it, because the sum of bounded sequences is bounded. $\endgroup$ – A. Salguero-Alarcón Feb 3 '17 at 22:53
1
$\begingroup$

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 $d((a_n)_n, (b_n)_n) = ||a_n - b_n||_\infty < \infty$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.