I have the following (possibly trivial) observation:
Let $K$ be an $\mathbb{F}$-vector space (I believe the argument also works for free modules), and let $X\subseteq K$ be it's basis with cardinality $\kappa:=\vert X\vert$. Denote also $\lambda:=\vert \mathbb{F}\vert$. Then:
$\vert K\vert=\begin{cases}
1 &, \kappa=0 \\
\lambda^\kappa &, \kappa,\lambda < \aleph_0 \\
\lambda &, 0<\kappa< \lambda, \lambda\geq \aleph_0 \\
\kappa &, \kappa\geq \vert X \vert
\end{cases}
$
For $\kappa \leq \aleph_0$ or I think the proof is relatively straight-forward, so I am not intersted. But when the cardinality is infinite the argument goes as follows (I write to verify). Define for all $m\in \mathbb{N}$
$K_m:=\Big\{ \sum\limits_{i=1}^m a_i \cdot x_i \Big\vert \; x_i\in X, a_i\in \mathbb{F} \quad \text{for all} \; i\in[m] \Big\}$
Then we know that $K=\underset{m=1}{\overset{\infty}{\bigcup}}K_m$, since $X$ generates $K$. We can see that:
$\vert K_m \vert =\vert \mathbb{F}\times X\vert^m= \big( \lambda \cdot \kappa \big)^m$
And since the cardinality of at least one is infinite, we can conclude that:
$\vert K_m\vert= \max\big\{ \lambda, \kappa \big\}$
And finally:
$\vert K\vert \leq \sum\limits_{m=1}^{\infty}\vert K_m\vert \leq \sum\limits_{m=1}^{\infty} \max\big\{ \lambda, \kappa \big\}= \max\big\{ \lambda, \kappa \big\}$
Finally since $K_m\subseteq K$, we know that $\max\big\{ \lambda, \kappa \big\}\leq \vert K\vert$. Which shows the observation.
First of all my question is (just to verify), does this observation not also translates to free modules?
Secondly, I can conclude by this arguement that $\vert K^*\vert=2^\kappa$ if $\kappa\geq\lambda$ and $\kappa\geq \aleph_0$. Is there an easy way to show that $\vert K^*\vert= 2^\kappa$ when $\lambda>\kappa \geq \aleph_0$?
I would appreciate any new insight into the matter.
Best Answer
If $|X|=\kappa$ and $\mathbb{F}=\lambda$, then:
The cardinality of $K$ is $\lambda^{(\kappa)}$, the cardinality of all functions $\kappa\to\lambda$ of finite support. To see this, note that each function $f\colon\kappa\to\lambda$ with finite support corresponds to the element $\sum_{x\in X}f(x)x$, where $f(x)$ represents the image of the element corresponding to $x$ in a fixed bijection between $X$ and $\kappa$.
The cardinality of $K^*$ is $\lambda^{\kappa}$, the cardinality of the set of all $\kappa\to\lambda$. To see this, note that the value of a functional is completely determined by its value on a basis, and the value on a basis is arbitrary. So given a function $g\colon X\to\mathbb{F}$, this defines a functional by extending linearly.
Now, if $\lambda\leq\kappa$ and $\kappa$ is infinite, then you have (since $\lambda$ is at least $2$): $$2^{\kappa}\leq \lambda^{\kappa}\leq \kappa^{\kappa} \leq (2^{\kappa})^{\kappa} = 2^{\kappa\kappa} = 2^{\kappa}.$$ This gives equality throughout, hence $|K^*| = \lambda^{\kappa} = 2^{\kappa}$.
(Note: this assumes the Axiom of Choice, since the statement that $\kappa\kappa=\kappa$ for all infinite cardinals $\kappa$ is equivalent to AC; not to mention existence of bases for vector spaces, and nontrivial duals).
See also this post which proves that if $\kappa$ is infinite, then the cardinality of $K^*$ is strictly larger than the cardinality of $K$, regardless of how $\kappa$ and $\lambda$ compare, so that the algebraic dual of an infinite dimensional space can never be isomorphic to the space itself.