Proof that two bases of a vector space have the same cardinality in the infinite-dimensional case.

Question

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?

Answer 1

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 linearly 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.

Answer 2

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 $|I|=|J|$.

Answer 3

Here's an approach in which the argument for the finite-dimensional case works in the infinite-dimensional case if you use Zorn's Lemma (which is equivalent to the axiom if choice).

Let $\mathsf k$ be a field. All vector spaces are assumed to be over $\mathsf k$.

Lemma 1: Let $V$ be a vector space and suppose that $S \subseteq V$ is any subset.

1). If $S$ is linearly dependent, then exists $T\subseteq S$ a finite subset together with scalars $\{\lambda_t \in \mathsf k^{\times}: t \in T\}$ such that $\sum_{t\in T}\lambda_t t=0$.

2). If $T$ is above and $t_0 \in T$ then we have $\text{span}_{\mathsf k}(S) = \text{span}_{\mathsf k}(S\backslash\{t_0\})$.

Proof: For $1)$, note that since $S$ is linearly dependent, there is a subset $T \subseteq S$ such that $\sum_{t\in T} \lambda_t t=0$, where $\lambda_t \in \mathsf k$ and not all $\lambda_t =0$. By replacing $T$ with $\{t \in T: \lambda_t\neq 0\}$ we may in fact assume that $\lambda_t \neq 0$ for all $t \in T$. For $2)$, note that if $t_0 \in T$ is any element of $T$ we have $$ t_0 = \sum_{t\neq t_0} (-\lambda_{t_0}^{-1}\lambda_t).t $$ and hence $\text{span}_{\mathsf k}(T\backslash\{t_0\}) = \text{span}_{\mathsf k}(T)$. It follows immediately that $\text{span}(S) = \text{span}_{\mathsf k}(S\backslash\{t_0\})$. $\phantom{asdfasdfasdfasdfasdfasdfasdfasdfasdfasdfasdf}\Box$

Now let $$ \mathcal I = \{I \subseteq V: I \text{ is linearly independent}\}, \quad \mathcal S = \{S\subseteq V: \text{span}_{\mathsf k}(S) = V\}. $$ Note that $\emptyset \in \mathcal I$ and $V \in \mathcal S$.

Proposition: If $I \in \mathcal I$ and $S \in \mathcal S$ then there is an injective map $f\colon I \to S$ such that $S\backslash\{f(I)\}\cup I \in \mathcal S$. In particular, for any $I\in \mathcal I$ and any $S \in \mathcal S$ we have the $|I|\leq |S|$, where $|X|$ denotes the cardinality of a set $X$.

Proof: Let $\mathcal C = \{(f,C):C\subseteq I, f\colon C\to S, f \text{ injective and } (S\backslash f(C))\cup C \in \mathcal S\}$. Then $\mathcal C$ is a poset with respect to the partial order $\preccurlyeq$ where $(f_1,C_1)\preccurlyeq (f_2,C_2)$ if $C_1 \subseteq C_2$ and $f_{2|C_1} = f_1$. By Lemma 2 below, $\mathcal C$ has a maximal element, $(f,C)$ say. To prove the proposition it suffices to show that $C=I$.

For the sake of a contradiction, suppose that $C\subsetneq I$. Let $S_1 = (S\backslash f(C))\cup C$. Then $S_1 \in \mathcal S$ since $(f,C) \in \mathcal C$. Now if $C \subsetneq I$ we may pick some $b \in I\backslash C$. Then $b \in \text{span}_{\mathsf k}(S_1)$, and hence there is a finite subset $T \subseteq S_1$ and scalars $\{\lambda_t: t \in T\}$ such that \begin{equation} \label{lindep} \tag{$\dagger$} b - \sum_{t \in T} \lambda_t t =0. \end{equation} Replacing $T$ by $\{t\in T: \lambda_t \neq 0\}$ we may assume that $(\dagger)$ is a linear dependence for $\tilde{T}:=T\cup \{b\}$ in which every element of $\tilde{T}$ appears with non-zero coefficient. Since $I$ is linearly independent, $b \notin \text{span}_{\mathsf k}(C)$ and hence we may pick $t_0 \in T\backslash C$. Let $\tilde{C} = C\cup \{b\}$ and $\tilde{S} = S_1\cup \{b\}$. Then by part 2) of Lemma 1 applied to the set $\tilde{S}$ with the dependence ($\dagger$) and each of the elements $b$ and $t_0$, we have $$ \text{span}_{\mathsf k}(S_1) = \text{span}_{\mathsf k}(\tilde{S}) = \text{span}_{\mathsf k}(\tilde{S}\backslash\{t_0\}) = \text{span}_{\mathsf k}(\tilde{C}\cup(S_1\backslash (C\cup\{t_0\}))). $$ It follows that if $g\colon \tilde{C}\to S$ be given by $g_{|C} = f$ and $g(b)=t_0$ then $g$ is injective and $\tilde{C}\cup(S\backslash g(\tilde{C}))= \tilde{S}\backslash\{t_0\}\in \mathcal S$ and hence $(g,\tilde{C}) \in \mathcal C$, contradicting the maximality of $(f,C)$. It follows that $C = I$ as required. $\phantom{asdfasdasdfasasdfasdfasdfdfasdffasdf}\Box$

Corollary Any $S \in \mathcal S$ contains a basis and any two bases have the same cardinality.

Proof: Let $S \in \mathcal S$ and let $\mathcal I_S = \{I \in \mathcal I: I \subseteq S\}$. We claim that any maximal element of $\mathcal I_S$ (with the containment partial order) is a basis of $V$. Indeed if $\text{span}_{\mathsf k}(I)\neq V$ then since $S$ spans $V$ there must be some $s \in S\backslash \text{span}_{\mathsf k}(I)$ so that $I\cup \{s\} \in \mathcal I_S$, contradicting the maximality of $I$. It follows that $I$ is a basis, and hence $S$ contains a basis of $V$. Since $\mathcal S$ is non-empty, every vector space has a basis. The fact that $\mathcal I_S$ always contains a maximal element is shown in Lemma 2 below.

Now by definition, $B$ is a basis of $V$ if $B \in \mathcal I \cap \mathcal S$. But it follows immediately from the previous proposition any two elements of $\mathcal I \cap \mathcal S$ have the same cardinality. Indeed if $B,B' \in \mathcal I\cap\mathcal S$ then since $B\in \mathcal I$ and $B' \in \mathcal S$, we have $|B|\leq |B'|$, and hence by symmetry we also have $|B'|\leq |B|$. It follows that $|B|=|B'|$. (If $V$ is finite-dimensional, then $\mathcal S$ contains a finite set and hence $|B|$ and $|B'|$ are finite and the implication $|B|=|B'|$ is trivial. If not, we must use the Schroder-Bernstein theorem to see that $|B|=|B'|$). $\phantom{asdfasdf}\Box$

Lemma 2: The partially ordered sets $\mathcal C$ and $\mathcal I_S$ (for $S\in \mathcal S$) have maximal elements.

Proof: We first show this for $\mathcal C$. If $S \in \mathcal S$ is finite, then if $(f,C) \in \mathcal C$, the cardinality of $C$ is bounded above by the cardinality of $S$. If $k$ is the maximum of $\{|C|: (f,C) \in\mathcal C\}$ then any $(f,C)\in \mathcal C$ with $|C|=k$ must be maximal with respect to the partial order $\preccurlyeq$. If $S$ is not finite then we use Zorn's Lemma, which asserts that any poset in which every chain has an upper bound has maximal elements, where a chain is an hence we must show that any chain in $\mathcal C$ (that is, any ordered subset of the poset $\mathcal C$) has an upper bound. Suppose that $\{(f_j,C_j):j \in J\}$ is such a subset of $\mathcal C$. If we set $C = \bigcup_{j \in J} C_j$ and let $f\colon C\to S$ be given by $f_{|C_j} = f_j$, then it is clear that $(C,f) \in \mathcal C$ and $(C,f)$ is an upper bound for the chain.

Similarly, if $V$ is finite-dimensional then $V$ has a finite spanning set $S$, and any $I\in \mathcal I_S$ has $|I|\leq |S|$ by the Proposition, and hence any $I\in \mathcal I_S$ of maximal cardinality will be maximal with respect to containment. Otherwise, we must again use Zorn's Lemma and hence must show that if $(I_j:j \in J)$ is a chain in $\mathcal I$ then it has an upper bound in $\mathcal I$. But if $K=\bigcup_{j \in J} I_j$ then $K\subseteq S$ and given a finite subset $D$ of $K$, say $D =\{d_1,\ldots, d_k\}$ we have $d_i \in I_{j_i}$ for some $\{j_i \in J: 1\leq i \leq k\}$. Since the sets $\{I_{j_i}: 1\leq i\leq k\}$ are all comparable, there is some $j_0 \in \{j_1,\ldots,j_k\}$ such that $I_{j_i}\subseteq I_{j_0}$ for all $i\in \{1,\ldots, k\}$ and hence $D\subseteq I_{j_0}$. Since $I_{j_0}$ is linearly independent, so is $D$, and since $D$ was arbitrary it follows that $K$ is linearly independent and hence $K \in \mathcal I_S$ is an upper bound for the chain $(I_j: j \in J)$, and so by Zorn's Lemma $\mathcal I_S$ has a maximal element. $\phantom{asdfasdfasdfasdf}\Box$