In $\textit{Advanced Linear Algebra}$ by Steven Roman, in order to prove $dim(V)<dim(V^*)$in infinite-dimensional vector spaces, he used the following inequality $$dim_K((K^\mathcal{B})_0)=dim_F((F^\mathcal{B})_0)<dim_F(F^\mathcal{B}) \le dim_K(K^\mathcal{B}),$$ where $F$ is a subfield of Field $K$, and $(A^\mathcal{B})_0$ is the set of all functions with finite support from $\mathcal{B}$ to $A$ and $A^\mathcal{B}$ is the set of all functions from $\mathcal{B}$ to $A$. $\mathcal{B}$ is also a basis for $V$.
Now, I am wondering what is the example of fields $K$ and $F$ such that the last inequality becomes strict, i.e. $dim_F(F^\mathcal{B}) < dim_K(K^\mathcal{B}).$
Equivalently, I can ask what is the example of a vector space $V$ such that if $dim(V)=card(X)$, where $X$ is a set with infinite elements, then $dim(V^*)>card(P(X))$, where $P(X)$ is the power set of $X$.
If $I$ is infinite, then $\dim_K(K^I)$ is always equal to the cardinality $|K^I|$ (see here). In particular, then, if $|K|>2^{|I|}$, then $\dim_K(K^I)=|K^I|\geq|K|>2^{|I|}$. So, any field of cardinality greater than $2^{|I|}$ gives a counterexample.