Prove that the cardinality of a vector space equals to the cardinality of its basis It is an exercise problem 11.1.12 in Dummit book.

If $F$ is a field with a finite or countable number of elements and $V$ is an infinite dimensional vector space over $F$ with basis $\mathcal{B}$, prove that the cardinality of $V$ equals the cardinality of $\mathcal{B}$.

So......
It is just clear that $|\mathcal{B}|\leq |V|$. So we only need to prove the reversed inequality. So I pick an element $v\in F$. Then it would be the linear combination of elements in $\mathcal{B}$.... And I am stuck.   I guess this way is not a good way to prove it. I hope to get some help from here! 
Thanks in advance!
 A: According to the tags, we are dealing with a Hamel basis, so all of our linear combinations are finite. For each $n \ge 1$, consider the map
$$E_n : F^n \times \mathcal{B}^n \to V : (a_1, \ldots, a_n, v_1, \ldots, v_n) \mapsto a_1 v_1 + \ldots + a_n v_n,$$
and define $E : \bigcup_n (F^n \times \mathcal{B}^n) \to V$ to be the union of the above functions (i.e. the set of all ordered pairs in each $E_n$, considered as a relation). As $\mathcal{B}$ spans $V$ in the Hamel sense, meaning every element of $V$ is a finite linear combination of elements of $\mathcal{B}$, we get that $E$ is surjective. Hence,
$$\left|\bigcup_n (F^n \times \mathcal{B}^n)\right| \ge |V|.$$
By assumption, we know that $|F| \le |\Bbb{N}| \le |\mathcal{B}|$. This means, there is an injection $\phi : F \to \mathcal{B}$. Note that,
$$\psi_n : F^n \times \mathcal{B}^n \to \mathcal{B}^n \times \mathcal{B}^n : (a_1, \ldots, a_n, v_1, \ldots, v_n) \mapsto (\phi(a_1), \ldots, \phi(a_n), v_1, \ldots, v_n)$$
is also an injection. Further, if $\psi$ is the union of the above $\psi_n$, then $\psi$ is an injective function from $\bigcup_n (F^n \times \mathcal{B}^n)$ to $\bigcup_n (\mathcal{B}^n \times \mathcal{B}^n)$, proving,
$$\left|\bigcup_n \mathcal{B}^{2n}\right| \ge \left|\bigcup_n (F^n \times \mathcal{B}^n)\right| \ge |V|.$$
But, as it turns out, given an infinite set $S$, $|S \times S| = |S|$. Hence, inductively, $|\mathcal{B}^m| = |\mathcal{B}|$ for all $m$. Therefore, there must exist bijections $\gamma_m : \mathcal{B}^m \to \{m/2\} \times \mathcal{B}$. If we let $\gamma$ be the union of $\gamma_m$ as $m$ ranges over positive even integers, then
$$\gamma : \bigcup_n \mathcal{B}^{2n} \to \bigcup_n (\{n \} \times \mathcal{B}) = \Bbb{N} \times \mathcal{B}$$
is a bijection. Thus,
$$|\Bbb{N} \times \mathcal{B}| = \left|\bigcup_n \mathcal{B}^{2n}\right| \ge \left|\bigcup_n (F^n \times \mathcal{B}^n)\right| \ge |V|.$$
Finally, since $|\Bbb{N}| \le |\mathcal{B}|$,
$$|\mathcal{B}| = |\mathcal{B} \times \mathcal{B}| \ge |\Bbb{N} \times \mathcal{B}| = \left|\bigcup_n \mathcal{B}^{2n}\right| \ge \left|\bigcup_n (F^n \times \mathcal{B}^n)\right| \ge |V|.$$
A: The conjecture is false as can be seen by a simple example using the field $F=(0,1)$ with addition mod$2$.  The vector space $V$ consists of all countable sequences of elements of $F$.  The basis $B=${$b_n$} where $b_n=1$ at the $n^{th}$ position and $=0$ otherwise.  $V$ is the equivalent to the set of subsets of $B$. The cardinality of $B$ is countable while that of $V$ is not.
