# 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)

• The only cases where this is not possible is when the field is finite or countably infinite. – egreg Jun 26 '15 at 13:35
• @egreg obviously yes there is no smaller cardinality than that of N. This is possible with CC also – Sushil Jun 26 '15 at 13:56

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.

• @Zev Chonoles Since lhf specified we were looking at the set of functions $X\to K$ which were nonzero for at most finitely many $x \in X$, the indicator functions do form a basis (though they would not if we looked at the larger space of all functions from $X$ to $K$ and $X$ were infinite). – Strants Jun 26 '15 at 4:38
• @Strants: Oops, I just missed that part it seems. – Zev Chonoles Jun 26 '15 at 4:39
• This is the smallest possible example, provided $\mathsf{CH}$ holds. – Matemáticos Chibchas Jun 26 '15 at 5:00
• @egreg why CH has no role here in proving it is smallest possible example. If CH fails, I'll take K to be set with cardinality between that of A and R. – Sushil Jun 26 '15 at 13:59
• No not in it. But when u said. "This is smallest possible example." – Sushil Jun 26 '15 at 15:36

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 $|X|<|K|$.