# 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!

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

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

• Thank you so much! Just for making sure, it should be $\phi(0)=0$ to make $\psi_n$ injective. Am I right? Jul 4, 2019 at 1:11
• @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! Jul 4, 2019 at 1:14
• Oh! I was stupid! Thank you! solution seems very nice! Jul 4, 2019 at 1:16
• @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. Jul 4, 2019 at 16:04
• @ 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. 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.

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