2
$\begingroup$

We have the following definition about convergence in a normed space:

"Let $(x_n)_{n=1}^\infty$ be a sequence in a normed space $(X,\|\cdot\|)$. We say that $x_n\to x$ in $X$ if,

$$d(x_n,x)\equiv \|x-x_n\|\to 0$$

as $n\to \infty.$"

My questions:

  1. How is it that we read this statement, and consequently understand it? In particular, what does it mean to talk about the norm (magnitude or length) of the difference of two sequences? Take the case that $x$ is some non-constant sequence $x_m$.

  2. I understand that a metric is induced by a norm, but how much of a distinction should be made between a norm and a metric within the context of a normed space? A normed space is automatically a metric space, and we talk about a metric $d$ being induced by the norm $d(x,y)=\|x-y\|$. Should I think of them, within the context of a normed space, as simply being the exact same, or should one take more care and try to distinguish them still?

$\endgroup$
  • 1
    $\begingroup$ I assume that the author was specializing from metric space to normed space. So it seems a little unnatural, but they're probably doing it to remind you that all of the results for metric spaces apply. There is economy of exposition in approaching the subject that way, but I can see how it would be confusing. $\endgroup$ – DisintegratingByParts Aug 12 '15 at 15:58
2
$\begingroup$
  1. I'm not sure I understand this question. The distance is defined to be the norm of the difference, and the definition is then the same as it is in metric spaces. You can get geometric intuition from Euclidean space, where $\| x - y \|$ is the length of the line segment connecting $x$ and $y$. If you replace $x$ with another sequence $y_m$, the situation is the same for each fixed $m$. This comes up for instance in talking about Cauchy sequences in metric spaces.
  2. They're not really exactly the same, but the analogy is close. $d(x,y)=\| x - y \|$ gives the metric from the norm; $\| x \| = \| x - 0 \| = d(x,0)$ gives the norm from the metric.
$\endgroup$

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.