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If $V$ is infinite dimensional with basis $A$, prove that $V^*$ is isomorphic to the direct product of copies of $F$ indexed by $A$. Deduce that dim $V^*$ > dim $V$.

I've shown in the past that the direct product of copies of $F$ indexed by $A$ is a vector space over $F$ and it has strictly larger dimension than the dimension of $V$. This fact would easily imply that dim $V^*$ > dim $V$ if I can show the isomorphism in the first part of the problem.

I think for the isomorphism, I have to find a bijection between the bases? I don't know how I would get a good map between $V^*$ and the direct product so that seems like a better idea. I'm having trouble getting off the ground with this one so any help would be appreciated.

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Given any $f\in V^*$, you can map $f\longmapsto \{f(a)\}_{a\in A}$. This map is clearly linear. It is one-to-one: if $f(a)=g(a)$ for all $a\in A$, then $f=g$. And it is onto: for any $\{\lambda_a\}_{a\in A}$, the map given by $f(a)=\lambda_a$ is mapped to $\{\lambda_a\}$. So that's your isomorphism.

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  • $\begingroup$ How is the set $\{f(a)\}_{a \in A}$ a subset of the direct product of copies of F? $\endgroup$ – user389056 Feb 7 '18 at 7:26
  • $\begingroup$ How do you define the direct product? $\endgroup$ – Martin Argerami Feb 7 '18 at 9:57
  • $\begingroup$ Does this set represent ordered pairs? Like, if $A$ was $\mathbb{N}$, then this would be $(f(1), f(2), \ldots )$? $\endgroup$ – user389056 Feb 7 '18 at 16:40
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    $\begingroup$ Yes, exactly. The right point of view, is that a sequence $a$ is a function $a:\mathbb N\to F$. You generalize that by defining the direct product $$\prod_{a\in A}F=\{\alpha|\ \alpha:A\to F\}.$$ $\endgroup$ – Martin Argerami Feb 7 '18 at 17:42

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