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I was thinking about the following question:

Can be possible to exist a normed vector space $(E,||\cdot||)$ (let us think all vector spaces are over $\mathbb{R})$ such that if $(F,||\cdot||_F)$ is another normed vector space then there is a linear map $T_F:F\to E$ that is an isometry onto its image, i.e. $||T_F(x)||=||x||_F$?

The idea is to think about a n.v.s that contains all the other n.v.s.

Jacaré
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  • Are you restricting $T_F(E) \subsetneq F$? – Rohan Didmishe Jan 02 '23 at 17:53
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    Well, one can take cartesian product of all normed spaces with only finite number of coordinates not null and then define norm as sum of norms on coordinates. – Salcio Jan 02 '23 at 18:00
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    Is $T_F$ supposed to be linear? If so, the answer is no and you can see this by taking $F$ to be a vector space whose dimension is a larger cardinal than the dimension of $E$. – Rhys Steele Jan 02 '23 at 18:01
  • @Salcio The problem with this is that normed spaces are a proper class, not a set. So there is no such thing as the cartesian product of all normed spaces. – Robert Israel Jan 02 '23 at 18:04
  • @RhysSteele Given a n.v.s. we always can find another n.v.s such that the cardinality of one is bigger than the other? – Jacaré Jan 02 '23 at 18:06
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    Yes, for any given cardinality you can take a set $X$ of that cardinality and define it to be a (Hamel) basis for the vector space $F$ (so that $F$ is the "free vector space over $X$"). This gives you a vector space of dimension $|X|$ which is enough since all vector spaces admit a norm (or for an explicit example of a norm on $F$; for $v \in F$, write $v = \sum_{i} \lambda_i x_i$ where $x_i \in X$ and only finitely many $\lambda_i$ are non-zero and define $|v| = \max_{i} |\lambda_i|$). – Rhys Steele Jan 02 '23 at 18:15
  • @RhysSteele I think this answer my question then. Thanks a lot. – Jacaré Jan 02 '23 at 18:19

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I'll convert my comments which show no such space $E$ exists to an answer.

Indeed, such an $E$ would have a Hamel basis $B$ and we could take a set $X$ of cardinality larger than $|B|$ and define $F$ to be the free vector space over $X$ so that $\dim F = |X| > |B| = \dim E$.

$F$ certainly admits a norm (e.g. for $v \in F$, write $v = \sum_{i} \lambda_i x_i$ where $x_i \in X$ and only finitely many $\lambda_i$ are non-zero and define $\|v\| = \max_{i} |\lambda_i|$). If $T_F: F \to E$ is an isometry then $T_F(X)$ is a basis for $T_F(F) \subset E$ and in particular is a linearly independent set in $E$ of cardinality larger than $|B|$ which is a contradiction.

Rhys Steele
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