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I am thinking about what makes $X^{**}$ not a reflexive space in infinite dimension case. I want to know what's wrong with the following proof:

Since every vector space $X$ has a basis $\{e_{\alpha}\}$.Then we can consider the basis $\{f_{\alpha}\}$ in $X^{*}$ s.t. $$f_{\alpha}(e_{\beta})=\delta_{\alpha\beta},$$ Then if we consider the basis $\{z_{\alpha}\}$ in $X^{**}$, we can define $z_{\alpha}$ s.t. $$z_{\alpha}(f_{\beta})=f_{\beta}(e_{\alpha})=\delta_{\alpha\beta}$$, which gives a natural isomorphic from $X\rightarrow X^{**}$.

89085731
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  • What do you mean by a reflexive space? Anyway, your $f_a$ does not form a basis, as they do not span the entire dual space. – Tobias Kildetoft Jun 25 '15 at 08:37
  • @TobiasKildetoft why it doesn't ? – 89085731 Jun 25 '15 at 08:40
  • Take the linear map that sends all basis elements to $1$. This map is not in the span of the $f_a$. – Tobias Kildetoft Jun 25 '15 at 08:41
  • @TobiasKildetoft it can be written as $\sum_{\alpha}f_{\alpha}$ – 89085731 Jun 25 '15 at 08:43
  • Try looking to see where your argument goes astray with the simple non-reflexive space $c_0$ of real sequences tending to $0$ with the supremum norm; its dual is $\ell_1$, and its second dual is $\ell_\infty$. – Brian M. Scott Jun 25 '15 at 08:44
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    That is an infinite sum, which is not well-defined. – Tobias Kildetoft Jun 25 '15 at 08:44
  • If $X$ is the set of real sequences with only finitely many terms, it is quite easy to identify $X^*$ to the set of all real sequences, which is strictly larger (and it's a general property of infinite-dimensional spaces, cf. for instance http://math.stackexchange.com/questions/1297845/the-relation-between-the-algebraic-dimensions-of-a-vector-space-and-its-dual/1299541#1299541). – PseudoNeo Jun 25 '15 at 08:44
  • @89085731: The span is the set of finite linear combinations. – Brian M. Scott Jun 25 '15 at 08:44
  • Of course, since $X^$ is in a sense always* strictly larger than $X$ when $\dim X$ is infinite, there's no hope for preserving a meaningful notion of reflexivity (at least purely algebraically, with topological vector space, the notion is highly interesting). – PseudoNeo Jun 25 '15 at 08:45
  • @BrianM.Scott: your first comment seems off-topic to me. The original question deals with "naked" vector spaces, without topology. – PseudoNeo Jun 25 '15 at 08:46
  • @BrianM.Scott I see. So what's way to form the basis in dual of vector space? – 89085731 Jun 25 '15 at 08:49
  • @PseudoNeo: I’ve not seen any definition of reflexive vector space that doesn’t require some topological structure. – Brian M. Scott Jun 25 '15 at 08:54
  • @BrianM.Scott: I guess some people use this word to talk about the purely algebraic fact that the canonical morphism $X \to X^{**}$ is an isomorphism if $X$ is a finite-dimensional vector space (Bourbaki does in the module setting, for an example), but I agree the vocabulary comes from the topological, infinite-dimensional case. – PseudoNeo Jun 25 '15 at 09:03

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