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 $\mathcal{B}$.

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!

  • $\begingroup$ What do you know about cardinality? Are there any results about cardinality that you think might be relevant? $\endgroup$ Jul 4, 2019 at 0:00
  • $\begingroup$ @TheoBendit Hmm... I am not quite sure. The definition of having same cardinality for two sets is having a bijection map between them. $\endgroup$
    – Lev Bahn
    Jul 4, 2019 at 0:01
  • $\begingroup$ @TheoBendit So I think it is enough to prove that there is an injection from $V$ to $\mathcal{B}$. But... does that exist? $\endgroup$
    – Lev Bahn
    Jul 4, 2019 at 0:03

2 Answers 2


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 $|\Bbb{N}| \le |\mathcal{B}|$, $$|\mathcal{B}| = |\mathcal{B} \times \mathcal{B}| \ge |\Bbb{N} \times \mathcal{B}| = \left|\bigcup_n \mathcal{B}^{2n}\right| \ge \left|\bigcup_n (F^n \times \mathcal{B}^n)\right| \ge |V|.$$

  • $\begingroup$ Thank you so much! Just for making sure, it should be $\phi(0)=0$ to make $\psi_n$ injective. Am I right? $\endgroup$
    – Lev Bahn
    Jul 4, 2019 at 1:11
  • $\begingroup$ @LevBan No, not necessarily. I'm guessing you're thinking of linear transformations here, and that injectivity of a linear transformation $T$ is equivalent to $Tx = 0$ only having the solution $x = 0$? In this case, we are not looking at linear transformations. I'm talking about an injection between unstructured sets here. Note that we definitely can't have $\phi(0) = 0$, as $0 \notin \mathcal{B}$, since $\mathcal{B}$ is linearly independent! $\endgroup$ Jul 4, 2019 at 1:14
  • $\begingroup$ Oh! I was stupid! Thank you! solution seems very nice! $\endgroup$
    – Lev Bahn
    Jul 4, 2019 at 1:16
  • $\begingroup$ @Theo Bendit The proof has a serious flaw in that it considers only finite linear combinations of elements of the basis. The vector space includes all combinations. $\endgroup$ Jul 4, 2019 at 16:04
  • $\begingroup$ @ Lev Ban The proof has a serious flaw in that it considers only finite linear combinations of elements of the basis. The vector space includes all combinations. $\endgroup$ Jul 4, 2019 at 16:06

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.

  • $\begingroup$ So.. it seems a way to prove $ \mathcal{B}\times F$ is countable. How is this related to the proof of my question? $\endgroup$
    – Lev Bahn
    Jul 4, 2019 at 0:10
  • $\begingroup$ You should check the definition of vector space. $\endgroup$ Dec 27, 2021 at 16:55
  • $\begingroup$ @Recently_registered I overlooked $F$ is countable. $\endgroup$ Dec 27, 2021 at 17:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.