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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!

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  • $\begingroup$ What do you know about cardinality? Are there any results about cardinality that you think might be relevant? $\endgroup$ – Theo Bendit Jul 4 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 Ban Jul 4 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 Ban Jul 4 at 0:03
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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 $|\Bbb{N}| \le |\mathcal{B}|$, $$|\mathcal{B}| = |\mathcal{B} \times \mathcal{B}| \ge |\Bbb{N} \times \mathcal{B}| = \left|\bigcup_n \mathcal{B}^{2n}\right| \ge \left|\bigcup_n (F^n \times \mathcal{B}^n)\right| \ge |V|.$$

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  • $\begingroup$ Thank you so much! Just for making sure, it should be $\phi(0)=0$ to make $\psi_n$ injective. Am I right? $\endgroup$ – Lev Ban Jul 4 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! $\endgroup$ – Theo Bendit Jul 4 at 1:14
  • $\begingroup$ Oh! I was stupid! Thank you! solution seems very nice! $\endgroup$ – Lev Ban Jul 4 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. $\endgroup$ – herb steinberg Jul 4 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. $\endgroup$ – herb steinberg Jul 4 at 16:06
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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.

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  • $\begingroup$ 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 Ban Jul 4 at 0:10

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