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I want an example where vector space $V$ is not isomorphic to it's double dual space.

I took space of sequences which have only finitely many non zero elements as $V$ and show that $V^*$ is a space of sequences, which has bigger dimension than $V$. Is there a quick way to show that $V$ and $V^{**}$ aren't isomorphic? Like injective function from $V^*$ to $V^{**}$ (possibly in general case)?

I know that there exist theorem which states that $\dim V \lt \dim V^*$ but I am looking for something easier.

José Carlos Santos
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romperextremeabuser
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    https://kconrad.math.uconn.edu/blurbs/linmultialg/dualspaceinfinite.pdf – Travis Willse Oct 19 '23 at 12:32
  • Why the tag functional analysis? Aren't you considering the algebraic dual and bidual? – Anne Bauval Oct 19 '23 at 12:32
  • I have already seen this paper and as I said I am looking for something easier – romperextremeabuser Oct 19 '23 at 12:35
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    @romperextremeabuser If so, then that kind of context is useful to include in your question. – Ben Grossmann Oct 19 '23 at 15:36
  • @romper The dual space of a vector space $V$ of countable dimension is easy to think about in terms of functions over some countable basis of $V$; this amounts to the observation that $V^$ will be isomorphic to a space of sequences, as you note. The problem is that the same trick does not apply to $V^{}$ because $V^$ doesn't have any nice basis; see this post for more on that. So, there's no analogous neat way to think of the elements of $V^{*}$ – Ben Grossmann Oct 19 '23 at 15:54
  • @romper You might have some luck thinking about the space of terminating sequences over $\Bbb F_2$. In this case, you can think of $V$ as the set of finite subsets of $\Bbb N$ and think of $V^$ as $\mathcal P(\Bbb N)$, the set of all subsets. $V^{*}$, I believe, can be thought of as the set of all functions from $f:\mathcal P(\Bbb N) \to \Bbb F_2$ that respect symmetric differences in the sense that $f(A \Delta B) = f(A) + f(B)$ – Ben Grossmann Oct 19 '23 at 15:59
  • Bounded sequences with sup-norm should do. – AlvinL Oct 20 '23 at 05:10
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    I think that one needs (some version of) the axiom of choice to produce elements of $V^{**}$ not belonging to (the canonical image of) $V$. This means, that the answer to your question is not as easy as you might hope. – Jochen Oct 20 '23 at 07:17

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