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The question of knowing whether is possible put a norm in a vector space is something new to me. I liked the Ivo ideas and I have wondered a little more about a particular case.

Is there a norm in which the vector space of all sequences with the induced metric is complete? This question sounds harder to answer. The completion of a metric space is something that is usually done to achieve this result. Is the completion of the norm given by Ivo ideas isomorphic to the vector space of all sequences? If not, is there a norm in which the vector space of all sequences with the induced metric is complete?

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    I think you will find the answer here https://math.stackexchange.com/questions/1234209/x-is-inner-product-space-then-its-completion-is-hilbert-space/1234330 (the question is about inner product spaces, but the answer covers normed spaces). – Rob Arthan Mar 25 '21 at 00:00

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By "Ivo ideas" I presume you mean, given a Hamel basis $B$ of vector space $V$, the supremum norm $\|v\| = \sup_{\beta \in B} |a_\beta|$ where $v = \sum_{\beta \in B} a_\beta \beta$.

An infinite-dimensional vector space $V$ with this norm is not complete. For example, let $\beta_i$, $i = 1,2,3,\ldots$, be a sequence of distinct members of $\beta$, and consider the sequence $v_n = \sum_{j=1}^n 2^{-j} \beta_{j}$. This is a Cauchy sequence, but it does not have a limit in this norm.

On the other hand, given a vector space $V$, to construct a norm in which it is complete all you need is to find a Banach space $B$ (i.e a complete normed linear space) with the same dimension (i.e. a Hamel basis of the same cardinality as that of $V$). Then the two spaces are isomorphic as vector spaces, and you just carry the norm of $B$ over to $V$. Specifically, if $\| \cdot \|_B$ is the norm of $B$, $(\alpha_i)_{i \in I}$ a Hamel basis of $V$ and $(\beta_i)_{i \in I}$ a Hamel basis of $B$ with the same index set, you define $$ \|v\|_V = \|\phi(v)\|_B$$ where if $v = \sum_i c_i \alpha_i$ is the representation for $v$ in the Hamel basis, $\phi(v) = \sum_i c_i \beta_i$.

Robert Israel
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    Your answer seems to suggest that the there is a Banach space with a Hamel basis of cardinality $\Bbb{N}$. I don't believe that is true. – Rob Arthan Mar 25 '21 at 00:26
  • Rob Arthan looks right: https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&ved=2ahUKEwjR_u3zocrvAhV2ILkGHfL3DzkQFjAAegQIAxAD&url=http%3A%2F%2Fuser.math.uzh.ch%2Fhalbeisen%2Fpublications%2Fpdf%2Fhamel.pdf&usg=AOvVaw0hJgpUZJs65l8VK7IGVyV4 – R. W. Prado Mar 25 '21 at 01:20
  • I did not mean to suggest that. If there is a Banach space with a Hamel basis of the same cardinality, there is a norm in which the given space is a Banach space. If there is not, then obviously there isn't. – Robert Israel Mar 25 '21 at 02:25
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    The OP's space of all (real) sequences has cardinality $\mathfrak c$ and a Hamel basis of that cardinality, and there are Banach spaces with Hamel basis of that cardinality. – Robert Israel Mar 25 '21 at 02:31
  • Aaah! Robert is right! A Hamel basis for the space of all sequences does not have Hamel basis of cardinality $\aleph_0$ but something equal or greater than $\mathfrak{c}$. That joke got me as well, but what Banach spaces is that, Robert? Your idea is pretty good. =) – R. W. Prado Mar 25 '21 at 03:00
  • Aaaah! Look that result https://www.jstor.org/stable/2318458 ! Wow, it's insane!

    As soon I absorb the ideas here, I will accept your answer. Hahaha If you have something to add, you can Robert! I appreciate any comment you have to do. =)

    – R. W. Prado Mar 25 '21 at 03:13
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To help people out, any separable Banach space have Hamel dimension of $\mathfrak{c}$, the second Beth number, as proven here or here. Since the Hamel dimension of all real sequences have the same cardinality $\mathfrak{c}$ and assuming the Continuum hypothesis, we can find a linear bijection $\phi$ from $l^{2}$ to the space of all sequences, say $\mathbb{R}^{\mathbb{N}}$. Hence, given a vector $w\in \mathbb{R}^{\mathbb{N}}$, define the norm $$\|w\|_{\mathbb{R}^{\mathbb{N}}} = \|\phi^{-1}(w)\|_{l^{2}}.$$ This norm define a complete Banach space. To define a complete Hilbert space it's just define a inner product in the same way.

  • I'm not sure if it's needed the Continuum hypothesis or not, but I feel that it is not needed and it does not sound a problematic hypothesis. If you have find it problematic, you can try post another answer here and I will be pleasured to like your answer. – R. W. Prado Mar 25 '21 at 18:59
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    you do need the continuous hypothesis to conclude both sets have the same cardinality. – L.F. Cavenaghi Apr 20 '21 at 01:40
  • In my proof, It was needed, but I'm quite convinced that it is not necessarily needed to prove my statement. The separable Banach space seems have some bijection with the space of all real sequences. For example, each vector in a separable Banach space can be seen as a sequence of natural numbers, since each vector is a limit of points in a enumerable set. – R. W. Prado Apr 20 '21 at 02:17
  • That's a "set theorist" matter of thing only. =) – R. W. Prado Apr 20 '21 at 02:29