linear algebra - Slick proof?: A vector space has the same dimension as its dual if and only if it is finite dimensional

It is clearly enough to show that an infinite dimensional vector space $V$ has smaller dimension that its dual $V^*$.

Let $B$ be a basis of $V$, let $mathcal P(B)$ be the set of its subsets, and for each $Ainmathcal P(B)$ let $chi_Ain V^*$ be the unique functional on $V$ such that the restriction $chi_A|_B$ is the characteristic function of $A$. This gives us a map $chi:Ainmathcal P(B)mapstochi_Ain V^*$.

Now a complete infinite boolean algebra $mathcal B$ contains an independent subset $X$ such that $|X|=|mathcal B|$---here, that $X$ be independent means that whenever $n,mgeq0$ and $x_1,dots,x_n,y_1,dots,y_min X$ we have $x_1cdots x_noverline y_1cdotsoverline y_nneq0$. (This is true in this generality according to [Balcar, B.; Franěk, F. Independent families in complete Boolean algebras. Trans. Amer. Math. Soc. 274 (1982), no. 2, 607--618. MR0675069], but when $mathcal B=mathcal P(Z)$ is the algebra of subsets of an infinite set $Z$, this is a classical theorem of [Fichtenholz, G. M; Kantorovich L. V. Sur les opérations linéaires dans l'espace des fonctions bornées. Studia Math. 5 (1934) 69--98.] and [Hausdorff, F. Über zwei Sätze von G. Fichtenholz und L. Kantorovich. Studia Math. 6 (1936) 18--19])

If $X$ is such an independent subset of $mathcal P(B)$ (which is a complete infinite boolean algebra), then $chi(X)$ is a linearly independent subset of $V^*$, as one can easily check. It follows that the dimension of $V^*$ is at least $|X|=|mathcal P(B)|$, which is strictly larger than $|B|$.

Later: The proof of the existence of an independent subset is not hard; it is given, for example, in this notes by J. D. Monk as Theorem 8.9. In any case, I think this proof is pretty because it captures precisely the intuition (or, rather, my intuition) of why this is true. I have not seen the paper by Fichtenhold and Kantorovich (I'd love to get a copy!) but judging from its title one sees that they were doing similar things...

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