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I am having a difficulty setting up the proof of the fact that two bases of a vector space have the same cardinality for the infinite-dimensional case. In particular, let $V$ be a vector space over a field $K$ and let $\left\{v_i\right\}_{i \in I}$ be a basis where $I$ is infinite countable. Let $\left\{u_j\right\}_{j \in J}$ be another basis. Then $J$ must be infinite countable as well. Any ideas on how to approach the proof?

user26857
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Manos
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2 Answers2

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In spirit, the proof is very similar to the proof that two finite bases must have the same cardinality: express each vector in one basis in terms of the vectors in the other basis, and leverage that to show the cardinalities must be equal, by using the fact that the "other" basis must span and be lineraly independent.

Suppose that $\{v_i\}_{i\in I}$ and $\{u_j\}_{j\in J}$ are two infinite bases for $V$.

For each $i\in I$, $v_i$ is in the linear span of $\{u_j\}_{j\in J}$. Therefore, there exists a finite subset $J_i\subseteq J$ such that $v_i$ is a linear combination of the vectors $\{u_j\}_{j\in J_i}$ (since a linear combination involves only finitely many vectors with nonzero coefficient).

Therefore, $V=\mathrm{span}(\{v_i\}_{i\in I}) \subseteq \mathrm{span}\{u_j\}_{j\in \cup J_i}$. Since no proper subset of $\{u_j\}_{j\in J}$ can span $V$, it follows that $J = \mathop{\cup}\limits_{i\in I}J_i$.

Now use this to show that $|J|\leq |I|$, and a symmetric argument to show that $|I|\leq |J|$.

Note. The argument I have in mind in the last line involves some (simple) cardinal arithmetic, but it is enough that at least some form of the Axiom of Choice may be needed in its full generality.

Arturo Magidin
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  • Just to check: Am I correct to say the dependence on the axiom of choice is contained entirely in the last line? It seems to me, though, that if $I$ and $J$ are both well-ordered then AC is not required. – Zhen Lin Jul 20 '11 at 17:25
  • @Zhen Lin: There is certainly some cardinal arithmetic going on in the last line, which would indeed invoke (at least in its most straighforward form) AC. Of course, even assuming vector spaces have bases is a nod in AC's direction... – Arturo Magidin Jul 20 '11 at 17:26
  • @Zhen, Arturo: Of course the assertion "Every vector space has a basis" is equivalent to the axiom of choice in its full power. Without it, however, there may be vector spaces with two bases of different cardinality. While this question requires the axiom of choice, note that if we assume that there are two bases to begin with then one could mistake this as a theorem of ZF, alas this is not the case. I have a feeling that it cannot be a theorem of ZF+"a proper fragment of AC" as well. Indeed, the universe of ZF is peculiar when consider ~AC :-) – Asaf Karagila Jul 20 '11 at 17:43
  • @Asaf: True; but if we have $I$ countable and $J$ infinite (as in the original question), then the above shows that $J$ is a countable union of finite sets (there might be some Choice involved in that). How much AC is needed to conclude that $J$ is countable in that case? (We can assume $J$ is not finite, since if $J$ is finite then every basis is finite of the same cardinality). – Arturo Magidin Jul 20 '11 at 18:04
  • @Arturo: If we assume that $J$ is a union of finite subsets of $I$ then no choice whatsoever. Otherwise I would say something about countable choice. (Note that a countable union of finite sets need not be countable!) – Asaf Karagila Jul 20 '11 at 18:08
  • @Asaf: We cannot assume $J$ is equal to a union of finite subsets of $I$ ($I$ and $J$ may be disjoint). For each $i\in I$, there exists $J_i\subseteq J$ which is finite, and such that $\cup J_i = J$. However, one may need Countable Choice to get the whole family ${J_i}_{i\in I}$ going; and of course, there's the fact about countable unions of finite sets. Does countable choice suffice to show a countable union of finite sets is countable? If so, for $I$ countable we can certainly get away with just countable AC. – Arturo Magidin Jul 20 '11 at 18:12
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    @Arturo: The axiom of countable choice is enough to show that a countable union of countable sets is countable (this can even be diminished to axiom of choice for countable family of countable sets, which is notably weaker!). I suppose you could get away even with axiom of choice for finite sets. – Asaf Karagila Jul 20 '11 at 18:14
  • @Arturo: Thank you for your answer. My understanding of cardinal arithmetic is rather limited so i am not sure how to proceed to show that $|J| \le |I|$. Can it be done in an elementary way, e.g. by constructing an injection of $J$ into $I$? Then by definition that would mean $|J| \le |I|$. – Manos Jul 21 '11 at 15:09
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    @Manos: Here's what I had in mind: Since $J=\cup J_i$, then $|J|=|\cup J_i| \leq \sum |J_i|$. Since each $J_i$ is finite, $|J_i|\leq\aleph_0$, so $|J|\leq \sum |J_i|\leq \sum \aleph_0$. Since the sum has $|I|$ many summands, $\sum \aleph_0 = |I|\aleph_0 = |I|$ (since $|I|$ is infinite). – Arturo Magidin Jul 21 '11 at 15:27
  • @Arturo: This is very instructive, thank you very much! – Manos Jul 21 '11 at 18:35
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    Somewhat late, but it turns out that BPI/Ultrafilter Lemma is enough to conclude that if $V$ has a basis, then all bases have the same cardinality. The proof is quite nice, too! – Asaf Karagila May 03 '12 at 21:00
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Once you have the necessary facts about infinite sets, the argument is very much like that used in the finite-dimensional case. The two crucial pieces of information are (1) that if $I$ is an infinite set of cardinality $\kappa$, say, then $I$ has $\kappa$ finite subsets, and (2) that if $|J|>\kappa$, and $J$ is expressed as the union of $\kappa$ subsets, then at least one of those subsets must be infinite.

Let $B_1 = \{v_i:i\in I \}$ and $B_2 = \{u_j:j \in J \}$, and suppose that $|J|>|I| = \kappa$. Each $u_j \in B_2$ can be written as a linear combination of some finite subset of $B_1$, say $u_j = \sum\limits_{i \in F_j}k_{ji}v_i$, where $F_j$ is a finite subset of $I$. For each finite $F \subseteq I$ let $J_F = \{j \in J:F_j = F\}$; clearly $J$ is the union of these sets $J_F$. But by (1) above $I$ has only $\kappa$ finite subsets, and $|J|>\kappa$, so by (2) above there must be some finite $F \subseteq I$ such that $J_F$ is infinite.

To simplify the notation, let $F = \{i_1,i_2,\dots,i_n\}$, and for $\mathcal{l}=1,2,\dots,n$ let $v_\mathcal{l} = v_{i_\mathcal{l}}$; then every vector $u_j$ with $j \in J_F$ is a linear combination of the vectors $v_1,v_2,\dots,v_n$. In other words, $\{u_j:j \in J_F\} \subseteq \operatorname{span}\{v_1,v_2,\dots,v_n\}$, and of course $\{u_j:j \in J_F\}$, being a subset of the basis $B_2$, is linearly independent. But $\operatorname{span}\{v_1,v_2,\dots,v_n\}$ is of dimension $n$ over $K$, so any set of more than $n$ vectors in $\operatorname{span}\{v_1,v_2,\dots,v_n\}$ must be linearly dependent, and we have a contradiction. It follows that we must have $|J| \le |I|$. By symmetry (or by the same argument with the rôles of $I$ and $J$ interchanged), $|I| \le |J|$, and hence $|I|=|J|$.

Brian M. Scott
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  • Thank you very much for your answer. Something that confuses me regarding cardinalities: if $I$ is infinite, there are two cases, right? Either $I$ is countable or uncountable. So when we say in (1) that $|I|=\kappa$, then $\kappa$ can actually have one of two values, the cardinality of the integers or the cardinality of the reals, right? – Manos Jul 21 '11 at 18:45
  • No, there are infinitely many different uncountable cardinalities. For example, the set of subsets of the reals has a larger cardinality than the set of reals itself. In fact, if $S$ is any set, $\mathscr{P}(S)$ has larger cardinality than $S$. – Brian M. Scott Jul 21 '11 at 19:25
  • @Brian To slightly shorten your proof, why not define from the start that $J_F={j\in J : u_j\in \text{span}{v_i : i \in F}}$, and then proceed exactly as above? – Aubrey Feb 02 '15 at 17:10
  • @Aubrey: Because the exposition is clearer the way I wrote it. – Brian M. Scott Feb 02 '15 at 19:03
  • @Brian M. Scott , What does it mean " $ k $ finite subset " , where $k$ is infinite cardinality? – hew Nov 07 '19 at 07:19
  • does it mean that $I$ is a disjoint union of finite subsets and the cardinality of disjoint sets is $k$ ? – hew Nov 07 '19 at 07:33
  • @hew: No, it means what I said: $I$ has cardinality $\kappa$, so it has only $\kappa$ different finite subsets. – Brian M. Scott Dec 23 '20 at 18:03