If two infinite vector spaces have the same cardinality then they are isomorphic Let $F$ be a field. Suppose that $V,W$ are $F$-vector spaces with infinite cardinality. Is it true that they have the same dimension?
If it is not true, will it be true if $|V|,|W| \geq |F|$?
 A: As they told you in the comments, that is not true even assuming $|V|=|W| \geq |F|$. Indeed, $|\mathbb{R}|=|\mathbb{R}^2|\geq |\mathbb{R}|$ but $\mathbb{R}$ and $\mathbb{R}^2$ are not isomorphic as $\mathbb{R}$-vector spaces.
But if you assume $|V|=|W| > |F|$ then it is true.
For every non-zero $F$-vector space $V$ we have that 
$$\max \{|F|, \mbox{dim}_{F}V \} \le |V| \le \max \{ \aleph_0, |F|, \mbox{dim}_{F}V \}$$
(see the answer in Cardinality of a vector space versus the cardinality of its basis). 
If $|V|=|W|=1$, the claim is clearly true. Thus we can assume $V$ and $W$ to be non-zero. If $F$ is finite, then either $|V|=|W|$ is at most countable and it is easy to show that this implies $\mbox{dim}_{F}V=\mbox{dim}_{F}W$ or $|V|=|W|$ is uncountable and then $\mbox{dim}_{F}V=|V|=|W|= \mbox{dim}_{F}W$. If $F$ is infinite, then $|F| \geq \mbox{dim}_{F}V$ cannot happen because we would have $|V|=|W|=|F|$. If $F$ is infinite and $|F| < \mbox{dim}_{F}V$, then $\mbox{dim}_{F}V=|V|=|W|= \mbox{dim}_{F}W$.
In either case we have $\mbox{dim}_{F}V= \mbox{dim}_{F}W$. So $V$ and $W$ are isomorphic.
