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$m$ is defined as the metric space of all limited real sequences and the book states that convergence in $m$ means convergence by coordinates, uniformly according to coordinate numbers. $s$ is defined as the metric space of all real sequences and the book states that convergence in $s$ means convergence by coordinates non-uniformly by coordinate numbers. I can't figure out what uniformly/non-uniformly means in this context. I thought it didn't change much, but now I notice, that $s$ is said that is a complete space because the convergence in it is convergence by coordinates, but $m$'s completeness isn't shown the same way, instead, there is a complete proof of it.

H-a-y-K
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1 Answers1

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"The" (usual) topology on your space $m$ (the vector space of real bounded sequences) is the topology of normed vector space associated to the norm $$\|(x_n)_n\|=\sup_{n\in\Bbb N}|x_n|.$$This way, a sequence $(x^{(k)})_k$ of bounded sequences $(x^{(k)}_n)_n$ converges to a bounded sequence $(x_n)_n$ iff $\lim_{k\to\infty}\left(\sup_{n\in\Bbb N}|x^{(k)}_n-x_n|\right)=0,$ i.e. $$\forall\epsilon>0\quad\exists K\quad\forall k\ge K\quad\forall n\quad|x^{(k)}_n-x_n|\le\epsilon$$ or equivalently: $$\forall\epsilon>0\quad\exists K\quad\forall n\quad\forall k\ge K\quad|x^{(k)}_n-x_n|\le\epsilon$$ This is what your book calls "convergence by coordinates, uniformly according to coordinate numbers".

It is a stronger requirement than "convergence by coordinates", which writes $$\forall\epsilon>0\quad\forall n\quad\exists K\quad\forall k\ge K\quad\quad|x^{(k)}_n-x_n|\le\epsilon.$$ In the latter, $K$ is allowed to vary with $n,$ whereas in the former it is not.

On your space $s=\Bbb R^{\Bbb N}$ (the vector space of all real sequences), "the" (usual) topology, corresponding to this weaker notion of convergence, is the product topology. A countable product of metric spaces is metrizable in various ways, some of which (the usual ones) preserve completeness.

Anne Bauval
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  • Thanks for your answer! – H-a-y-K Oct 26 '22 at 08:08
  • I have a question though. So uniform convergence by coordinates is a stronger version of convergence by coordinates, right? If convergence in a space means convergence by coordinates then that space is complete. But why can't we directly deduce the same from the fact that convergence in $m$ means uniform convergence by coordinates? In other words, if convergence in $m$ is stronger than convergence in $s$ and $s$ is complete because convergence in $s$ is convergence by coordinates, then why can't we deduce the same way that $m$ is complete? – H-a-y-K Oct 26 '22 at 08:10
  • Completeness is not a topological property but a metric one (or more generally a uniform one, but forget). 2. Not every countable product of metric spaces $(E_n,d_n)$, even equipped with one of the "usual" metrics associated to the $d_n$'s, is complete: a necessary condition is that the $(E_n,d_n)$'s are complete. 3. If you have 2 metrics $d$ and $d'$ on a space and if convergence for $d$ implies convergence for $d'$, completeness for $d'$ does not imply completeness for $d$ because even if a $d$-Cauchy sequence is $d'$-Cauchy hence $d'$-convergent, it may be non-$d$-convergent.
    • – Anne Bauval Oct 26 '22 at 09:38