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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!

Lev Bahn
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  • What do you know about cardinality? Are there any results about cardinality that you think might be relevant? – Theo Bendit Jul 04 '19 at 00:00
  • @TheoBendit Hmm... I am not quite sure. The definition of having same cardinality for two sets is having a bijection map between them. – Lev Bahn Jul 04 '19 at 00:01
  • @TheoBendit So I think it is enough to prove that there is an injection from $V$ to $\mathcal{B}$. But... does that exist? – Lev Bahn Jul 04 '19 at 00:03

2 Answers2

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

Theo Bendit
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  • Thank you so much! Just for making sure, it should be $\phi(0)=0$ to make $\psi_n$ injective. Am I right? – Lev Bahn Jul 04 '19 at 01:11
  • @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! – Theo Bendit Jul 04 '19 at 01:14
  • Oh! I was stupid! Thank you! solution seems very nice! – Lev Bahn Jul 04 '19 at 01:16
  • @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. – herb steinberg Jul 04 '19 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. – herb steinberg Jul 04 '19 at 16:06
  • @herbsteinberg This is not a flaw; it's part of the definition of Hamel basis (see the tags on the question). For that matter, countable combinations don't make sense in arbitrary vector spaces over arbitrary fields. You'll need topologies on your field and vector space before you can define countable combinations. – Theo Bendit Jul 04 '19 at 22:17
  • @Theo Bendit It seems that the original question is ambiguous. Given a basis, the smallest vector space defined by it requires only finite combinations of basis elements. However a vector space using all combinations is definable and obviously bigger. – herb steinberg Jul 06 '19 at 16:50
  • @herbsteinberg While I agree the asker could have made the type of basis more explicit, I don't see the ambiguity. I'm not aware of any definition for countable sums in abstract vector spaces. How would you define them? For example, in the finite vector space $\Bbb{F}_2$ over the same field, then what would be, say, $1 + 0 + 1 + 0 + 1 + 0 + \ldots$? Or how about $1 + 1 + 1 + 1 + \ldots$? – Theo Bendit Jul 08 '19 at 02:59
  • @Theo Bendit I am slightly confused by your terminology. In any case, look at the answer I gave to the original question. How would a basis be defined for the vector space I described - two element scalar field and vector space consisting of all possible sequences? All this with no topology. – herb steinberg Jul 08 '19 at 17:30
  • @herbsteinberg I doubt there would be a basis that could be constructed in ZF; some form of axiom of choice would probably be necessary. That is, I suspect it is not possible to give an explicit description of such a basis. But, Hamel bases exist for every vector space under ZFC. This is done by a standard Zorn's lemma argument on the partial order of linearly independent sets, ordered by set inclusion (you can see the wiki page for more details). – Theo Bendit Jul 09 '19 at 00:23
  • @Theo Bendit Thank you for your response. My education concerning vector spaces was more about normed linear spaces, where a basis, particularly for $l_p$ or $L_p$ was readily definable. Here the finite linear combination axiom did not apply. – herb steinberg Jul 09 '19 at 01:41
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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.

herb steinberg
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