1
$\begingroup$

I am slightly confused about one aspect of the definition of a convergent sequence on a normed vector space. The definition states that a sequence $\{x_n\}$ in the normed space $(X,||\;||)$ is convergent if $$||x_n-x||\le \varepsilon \;\forall n \gt N$$ where $N$ is some natural number and $x$ is called the limit of convergence. Now every source I have looked at says that $x$ needs to be in $X$ - is this the standard? i.e. would we not call a sequence that converges to an $x \notin X$ convergent?

(A source would be helpful)

$\endgroup$
4
  • $\begingroup$ where does this $x$ live then? $\endgroup$ Commented Jul 31, 2017 at 14:47
  • $\begingroup$ @JensRenders I have read it is possible for $x$ to live in e.g. a hole that is not part of $X$. One example is if $X$ is the set of rational numbers the sequence might tend to an irrational number (Keener, 1995). $\endgroup$ Commented Jul 31, 2017 at 14:48
  • $\begingroup$ That is true, then X is a subspace of a bigger space Y (in this case your bigger space is $\mathbb{R}$) see my answer $\endgroup$ Commented Jul 31, 2017 at 14:53
  • $\begingroup$ Is this definition of convergence just completely equivalent to saying that: convergence in a normed space is convergence in the metric space induced by the norm; or am I misunderstanding something? $\endgroup$ Commented Oct 23, 2019 at 11:28

1 Answer 1

2
$\begingroup$

$x$ must be an element of $X$ to ensure that $x_n-x$ has meaning and that this is again an element of $X$ so we can look at its norm: $\lVert x_n-x\rVert$

However, if the space $X$ is a subspace of a normed space $Y$ then it is possible for the limit to be outside of $X$. We then say that he sequence doesn't converge in $X$ but it does converge in $Y$.

A subspace $X$ for which every convergent sequence has it's limit still in $X$ is called a closed subspace

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .