An example for infinite dimensional vector space with Hamel dimension smaller than $\operatorname{card} F$ What will be the example for a vector space(infinite dimensional) over a field where Hamel basis has strictly smaller cardinality than that of field?
It is not possible in a Hilbert Space (over R or C)
 A: Here is kind of a silly way of constructing such objects.
Let $F$ be any field, and let $X$ be any infinite set. Then we can form the free $F$-vector space $V$ on $X$, consisting of all formal sums of finitely many terms
$$V=\{\textstyle\sum \alpha_ix_i:\alpha_i\in F,\;x_i\in X\}$$
of which $X$ is an obvious basis, so that $V$ has dimension $|X|$ as an $F$-vector space.
Now let $Y$ be any infinite set of cardinality strictly larger than that of $X$, and form the field of rational functions $K=F(Y)$, so that $|K|\geq |Y|>|X|$.
Then $V\otimes_F K$ has dimension $|X|$ as a $K$-vector space, but it is now a vector space over the field $K$ with $|X|<|K|$.
A: The space of $\mathbb R[X]$ of polynomials over $\mathbb R$ has a countable basis: $1, X, X^2, \dots$.
More generally, if $K$ is a field and $X$ is any set, then the space of all functions $X \to K$ having finite support is a $K$-vector space having a basis of cardinality $|X|$: a basis is given by the functions that are zero everywhere except at exactly one point of $X$. So, you only need to take $X$ an infinite set of cardinality strictly smaller than that of $K$. The example above is this one with $K=\mathbb R$ and $X=\mathbb N$. It is actually the smallest possible example.
