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Part of the proof that $\dim(V^*_F) > \dim(V_F)$ for an infinite dimensional space is that $\dim(\operatorname{Hom}(V_F, F)) \ge |F|$ (i.e $\dim(V^*_F) \ge |F|$). See for example Dual space question and some of its references.

It appears that for any space $W_F$ of non-zero dimension, with $\dim(V_F)$ still being infinite, that one should be able to embed $Hom(V_F, F)$ as a subspace of $\operatorname{Hom}(V_F, W_F)$ and therefore $\dim(\operatorname{Hom}(V_F, W_F)) \ge \dim(V^*_F) \ge |F|$.

I haven't seen this result anywhere: it seems obvious but have I made a mistake ?

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Tom Collinge
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Let $y \in W_F-\{0\}$, we have the embedding: $\phi \in Hom(V_F,F) \mapsto g_{\phi} \in Hom(V_F,W_F)$ where $g_{\phi}(x)=\phi(x)y$ for all $x \in V_F$. It's an embedding because $g_{\phi}=g_{\psi}$ implies $\forall x, \phi(x)y=\psi(x)y$. So, $\forall x \in V_F, (\phi(x)-\psi(x))y=0$. And $y \neq 0 \implies \phi(x)=\psi(x)$ for all $x$. So $\phi=\psi$.

So, the result is true.

marco2013
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