Let $X$ be the vector space of all real sequences . Does there exist a norm on $X$ which makes it complete ?

  • $\begingroup$ $\ell^p$ for any $p$ is a complete metric space, but this requires taking the subspace of vectors for which the norm is not infinite. Is that what you are looking for? $\endgroup$ – Eric Naslund Apr 19 '16 at 16:09
  • 6
    $\begingroup$ Well, if you find any Banach space with the same algebraic dimension than $X = \Bbb {R}^\Bbb {N} $, you can use any isomorphism between the two vector spaces to transfer the norm. The result will be complete. IIRC then every separable infinite dimensional Banach space has as algebraic dimension the cardinality of the continuum, as should $X $ (did not think too much about it). $\endgroup$ – PhoemueX Apr 19 '16 at 16:16
  • $\begingroup$ @PhoemueX is right. Your $X$ has algebraic dimension (the size of a basis an an abstract vector space over $\mathbb R$) equal to the cardinality of the continuum, so it is isomorphic (again as an abstract vector space) to any other real vector space of the same algebraic dimension, for example $l^2$. Pick an isomorphism and use it to transport to $X$ the usual norm on $l^2$. (This argument uses the axiom of choice to get bases and so make sense of "abstract dimension".) $\endgroup$ – Andreas Blass Apr 19 '16 at 16:21
  • $\begingroup$ @PhoemueX I did think some about why those dimensions are $c$ and decided I was missing a trick. Asked the Banach space guy at the office. See the Main Lemma in the latest version of my answer if you're still curious. $\endgroup$ – David C. Ullrich Apr 19 '16 at 22:42
  • $\begingroup$ @David: Thank you very much (also to the Banach space guy :). Sadly, I can't upvote your answer twice. $\endgroup$ – PhoemueX Apr 20 '16 at 6:23

The answer to the question exactly as you asked it is yes; your space is isomorphic as a vector space, with no topology, to various Banach spaces. (See various comments for details.)

Edit: The assertion that the answer is yes has met with vigorous disbelief. Also there's a technical point that I realized after some thought I simply didn't know how to do. I've added an Appendix at the bottom of this answer explaining things in detail, in particular showing that $\Bbb R^{\Bbb N}$ and $\ell^2(\Bbb N)$ do in fact have algebraic dimension $c$. Back to the short version:

But that norm simply has nothing to do with the structure of the space as a space of sequences. There's no complete norm on the space $X$ of all real sequences that has the sort of properties you must have in mind. In particular: For $n\in\Bbb N$ define $\Lambda_n:X\to\Bbb R$ by $$\Lambda_n x=x_n.$$There is no complete norm on $X$ such that every $\Lambda_n$ is bounded. (That is, there is no complete norm such that for every $n$, $x_n$ depends continuously on $x$.) In fact there's no such norm, complete or not.

Suppose on the other hand that $||\Lambda_n||=c_n$. Define $$S_N=\sum_{n=1}^N 2^{-n}c_n^{-1}\Lambda_n.$$Then $||S_N||<1$ for every $N$, which says that for every $x$ we have $$\left|\sum_{n=1}^N2^{-n}c_n^{-1}x_n\right|\le||x||$$for every $N$. This is impossible; for any sequence $c_n$ there exists a sequence $x$ such that $$\sup_N\left|\sum_{n=1}^N2^{-n}c_n^{-1}x_n\right|=\infty.$$

Appendix: Two things: why those bases show that the answer to the question is yes, and why those bases exist. Note that here "basis" means basis in the pure linear algebra sense, often called "Hamel basis": If $B$ is a basis for $X$ then every $x\in X$ is equal to a unique linear combination of finitely many elements of $B$. $\newcommand\S{\Bbb R^{\Bbb N}}$ $\newcommand\L{\ell^2(\Bbb N)}$

Assume for now that $\S$ and $\L$ both have bases of cardinality $c$. Then there is a bijection betweenn the two bases. This gives us an isomorphism $I:\S\to\L$, defined by mapping linear combinations of the elements of the basis for $\S$ to the corresponding linear combinations of the elements of the basis for $\L$. We can thus define a norm on $\S$ by $||x||_{\S}=||Ix||_{\L}$, and it is straightforward to verify that the new norm on $\S$ is complete (given a Cauchy sequence in $\S$, by definition the corresponding sequence in $\L$ is Cauchy, hence convergent, so that by definition the original sequence in $\S$ is convergent.)

Ok, I left out some details there. Every detail is trivial - this equals that be definition since this or that is an isomorphism.

How do we know there are bases of cardinality $c$? Both spaces have cardinality $c$ so a basis can be no larger than $c$; we need $c$ independent vectors. It seems clear that we somehow should get this from the fact that there are $c$ subsets of $\Bbb N$, but I didn't see how to get the independence. I asked the Banach-space guy at the office:

Main Lemma There exists a map $S:\Bbb R\to\mathcal P(\Bbb N)$ such that $S(r)$ is infinite for every $r$, and such that given finitely many distinct reals $r_1,\dots,r_n$ there exists $N$ so that the sets $S(r_1)\cap[N,\infty),\dots,S(r_n)\cap[N,\infty)$ are pairwise disjoint.

Proof Let $\Bbb Q=\{q_1,\dots\}$. For each $r\in\Bbb R$ choose a sequence $(q_{n_j})$ of distinct rationals with $q_{n_j}\to r$, and let $S(r)=\{n_j\}$. QED.

Now it's clear that $\{\chi_{S(r)}:r\in\Bbb R\}$ is a linearly independent subset of $\S$. And if $(e_1,\dots)$ is orthonormal in $\L$ then $\left\{\sum_{j\in S(r)}\frac{e_j}{j}:r\in R\right\}$ is a linearly independent subset of $\L$.


As a complement to the earlier (good) answer and comments: the space of all sequences (whether real or complex) arises in at least one fairly natural way, namely, as the continuous dual to the LF-space (strict inductive limit of Frechet spaces) $\mathbb R^\infty=\bigcup_n \mathbb R^n$, where $\mathbb R^n$ has its usual topology and is included in $\mathbb R^{n+1}$ on the first $n$ coordinates. Since the very definition of the thing expresses it as a countable union of nowhere dense sets, by Baire $\mathbb R^\infty$ cannot be complete-metrizable (with its LF-space topology). Its continuous dual is the collection $X$ of all sequences, with the projective-limit topology given by $X\to \mathbb R^n$ by taking the initial segment of length $n$. The latter "projection" maps are compatible... so $X$ is a Frechet space (with topology given by the countably-many seminorms that are any norms on the limitands $\mathbb R^n$). Its continuous dual is (takes a little work...) the LF-space $\mathbb R^\infty$ again, which is the wrong cardinality to be the dual of a Banach space, for example.

  • $\begingroup$ Quick note: by "is the wrong cardinality" you really mean "has basis of the wrong cardinality". $\endgroup$ – Jason Apr 19 '16 at 17:36
  • 2
    $\begingroup$ Also, I guess a person should say that this "answer" is mostly unrelated to issues about "algebraic/Hamel" bases, axiom of choice, etc. The point was to argue that a natural structure (complete metrizable topological vector spaces) is Frechet, but not Banach. $\endgroup$ – paul garrett Apr 19 '16 at 22:52

Your Answer

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