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I'm following the proof in http://www.science.oregonstate.edu/~parks/sample338.pdf .

Theorem 5 If V is infinite dimensional, then V and $V^∗$ are not isomorphic.

Proof $B$ basis of $V$, $\kappa=|B|$. Case I $2^{\aleph_0}\leq \kappa$. For each subset $S\subset B$ we can define a distinct element $$ \sum_{b\in S}b^*. $$ Thus we see that the cardinality of $V^∗$ is at least $2^\kappa$ which is strictly greater than $\kappa = \kappa\cdot 2^{\aleph_0}$, the cardinality of V. Case II not relevant for the question.

What is the relation between $|B|$ and $|V|$?

I understand that $|B|=\kappa\geq 2^{\aleph_0}$ is given, and that $|V^*|\geq 2^\kappa$. Additionally I have the identity for infinite $V$ $$ |V|=\textrm{max}(|\mathbb{F}|,|B|), $$ where $\mathbb{F}$ is the field under the vector space. Am I correct if I say that the author makes the assumption $|\mathbb{F}|\leq 2^{\aleph_0}$?

And another question (to understand where $\kappa = \kappa\cdot 2^{\aleph_0}$ comes from): Is $|V|=\textrm{max}(|\mathbb{F}|,|B|)$ equivalent to $|V|=|\mathbb{F}|\cdot|B|)$?

PeptideChain
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1 Answers1

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Twice yes: the first line of the linked note says we are considering vector spaces over the real numbers. So $|\mathbb{F}| = 2^{\aleph_0} = \mathfrak{c}$

And for all infinite cardinals $\kappa, \lambda$ $$\kappa \cdot \lambda = \kappa + \lambda = \max(\kappa, \lambda)$$

(at least if we assume the axiom of choice (AC), which we do, because we assume a vector space always has a basis)

So for all vector spaces over an infinite field $|V| = |\mathbb{F}| \cdot |B|$ (because we use finite linear combinations from $B$ to get all of $V$), so over the reals : $|V| = \mathfrak{c} \cdot |B|$ and so in case (i), when $|B| \ge 2^{\aleph_0}$: $|V| = |B|$, while the argument shows the first inequality in $$|V^\ast| \ge 2^{|B|} > |B| = |V|$$ the last inequality by Cantor's theorem.

Henno Brandsma
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  • If you know a book with the general prove would be nice. I found only Roman, Advanced Linear Algebra, 2008. Perhaps I try again with his proof. – PeptideChain May 25 '18 at 11:59
  • @LiPo the link you gave has the general proof. All basis-cardinalities have been done. In what sense do you need a more general proof? What is missing? – Henno Brandsma May 25 '18 at 12:17
  • I mean general in respect to the cardinality of the field. $2^{\aleph_0}$ is not the the maximal cardinality for a field, is it? – PeptideChain May 25 '18 at 12:20
  • here and here are other more extensive proofs on this site, for any field size. And by the (upward) Löwenheim-Skolem theorem, for every infinte cardinal $\kappa$ there is a field of that size. – Henno Brandsma May 25 '18 at 12:40