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Suppose that $V$ is a vector space over the field $\mathbb{K}$ with the norm $||.||$. Also suppose that $V$ is an incomplete normed space. What can be said about the cardinality of the basis of $V$? It's well-known that if $V$ is a Banach space (complete normed space), then the basis of $V$ has uncountable cardinality but what can be said about the incomplete case? Is there any general result?

Update: After reading the mentioned question and the helpful comments of David Gao, I wrote an answer. I don't know whether this question really counts as a duplicate, but anyway I close the question.

S.H.W
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  • You can find an example of a normed space with countable Hamel dimension. Try it! – GEdgar Oct 11 '24 at 17:34
  • @GEdgar Thanks I will try. If I find such an example what's the next step? I mean given such an example it's likely that every incomplete normed space has a countable basis. How to prove that? – S.H.W Oct 11 '24 at 17:59
  • @MartinBrandenburg Thanks. Would you explain how that post answers my question? My question is about the incomplete normed spaces. – S.H.W Oct 11 '24 at 18:03
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    Literally every infinite-dimensional space (over real or complex numbers) can be given a norm that makes it an incomplete normed vector space, for example the norm defined in the duplicate question Martin links to. So no, not every incomplete normed space has a countable basis. – David Gao Oct 11 '24 at 18:03
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    The norm in the question Martin links to makes the space into $c_{00}(I)$, the space of finitely supported functions $I \to \mathbb{K}$, where $I$ is the basis, equipped with the supremum norm. Unless $I$ is finite, this is always incomplete, which is an easy exercise to prove. – David Gao Oct 11 '24 at 18:06
  • @DavidGao Thanks. So it seems that an incomplete normed spaces can have countable or uncountable basis. Is this conclusion correct? – S.H.W Oct 11 '24 at 18:08
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    @S.H.W Yes. Its dimension can be any infinite cardinal. – David Gao Oct 11 '24 at 18:10
  • @DavidGao I'm trying to show that the mentioned normed space is incomplete but I couldn't find a Cauchy sequence which converges to ${\bf x} \not \in X$. According to what I have understood from the question, the norm is defined as $|{\bf x}| = \max{|a_i| \mid i \in F }$ where $a_i$'s are the coefficients of the $\bf x$ when $\bf x$ is expressed in the basis $({\bf e}i){i \in I}$. Would you give some hint, please? – S.H.W Oct 11 '24 at 19:53
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    @S.H.W Choose a countably infinite subset ${i_n}n$ of the index set $I$. Then consider the sequence $x_m = \sum{k=1}^m \frac{1}{k}e_{i_k}$. – David Gao Oct 11 '24 at 21:15
  • @DavidGao Thanks for the comments, I wrote an answer. I would appreciate your feedback on the answer. – S.H.W Oct 12 '24 at 11:14

1 Answers1

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Assuming the Axiom of Choice, suppose that the arbitrary vector space $V$ has the Hamel basis ${\cal B} = ({\bf e}_i)_{i \in I}$. Because ${\cal B}$ is a basis, the set ${\cal B}$ is a subset of $V$. So, $$\forall i\in I: {\bf e}_i\in V,$$which implies every finite linear combination of ${\bf e}_i$'s belongs to $V$. This means that any eventually-zero sequence $x_n = (a_1,a_2,a_3,\dots,a_n,0,0,0,\dots)$ corresponds to a vector in $V$. Conversely, every vector in $V$ corresponds to a eventually-zero sequence since by the definition of basis, every vector can be written as a finite linear combination of ${\bf e}_i$'s. Now define the norm $\|{\bf x}\| = \max\{|a_i| \mid i \in F \}$ where ${\bf x} =\sum_{i \in F}a_i{\bf e}_i$. It can be proved that $\|{\bf x}\|$ is indeed a well-defined norm. In order to prove that vector space $V$ with this norm, in the case $\dim(V)=\infty$, is an incomplete normed space, consider the sequence, $${\bf x}_n = \sum_{i=1}^n\frac{1}{i}{\bf e}_i \ \ \text{corresponds to} \ \ (\frac{1}{1},\frac{1}{2},\frac{1}{3},\dots,\frac{1}{n},0,0,0,\dots).$$ This sequence is Cauchy since if $m\gt n$ then, $$||{\bf x}_m - {\bf x}_n|| = ||\sum_{i=n+1}^m\frac{1}{i}{\bf e}_i|| = \frac{1}{n+1},$$ and $\frac{1}{n+1}$ tends to $0$ as $n$ and $m$ tend to $+\infty$ but the sequence is not convergent since it has to converge to some ${\bf x} =(a_1,a_2,\dots,a_m,0,0,\dots)$ and for $n \ge m+1$ we have, $$||{\bf x} - {\bf x}_n||\ge \frac{1}{m+1},$$which shows that the sequence is at a positive distance from any element in $V$. Interestingly, it can be shown that the space of eventually-zero sequences is not a complete space under any norm.

Conclusion: In the case $\dim(V) \lt \infty$, the vector space is always complete. If $\dim(V) = \infty$ and $V$ is an incomplete normed vector space, then the basis can be countable or uncountable but if $V$ is a complete normed vector space, then the basis is uncountable.

S.H.W
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