A vector space is said to be infinite dimensional iff it cannot be spanned by a finite set of vectors. Unlike the finite dimensional cases, not much is said about the dimension of infinite dimensional spaces in typical linear algebra courses. Therefore we supplement the analogous facts here, including that the dimension of an infinite dimensional space is well-defined. The following are some examples of infinite dimensional vector space:
Example.
(a) The set of real polynomials in one variable $x$ \[ \mathbb{R}[x] = \{a_n x^n + \dots + a_1 x + a_0 | n \in \mathbb{N} \text{ and } a_i \in \mathbb{R} \text{ for all } i \in \{0, ..., n\}\} \] as a real vector space has a countable basis, the set of monomials together with the constant polynomial 1, $\{1, x, x^2, x^3, \dots\}$.
(a) The set of real polynomials in one variable $x$ \[ \mathbb{R}[x] = \{a_n x^n + \dots + a_1 x + a_0 | n \in \mathbb{N} \text{ and } a_i \in \mathbb{R} \text{ for all } i \in \{0, ..., n\}\} \] as a real vector space has a countable basis, the set of monomials together with the constant polynomial 1, $\{1, x, x^2, x^3, \dots\}$.
(b) The set of infinite $\mathbb{F}_2$ sequences \[\mathbb{F}_2^\mathbb{N}\ = \{(b_0, b_1, b_2, \dots)|b_i \in \mathbb{F}_2 \text{ for all }i \in \mathbb{N}\}\]as an $\mathbb{F}_2$ vector space cannot be spanned by a countable subset. To see this, notice that the vector space is uncountable but the span of a countable subset using a finite field is countable. \[\operatorname{span}_{\mathbb{F}_2}(\{v_1, v_2, \dots\}) = \cup_{k=1}^{\infty}{\{\sum_{i=1}^{k}{c_i v_i}}|c_i \in \mathbb{F}_2 \text{ for all } i \in \{1,2,\dots,k\}\}.\]
(c) The set of infinite real number sequences \[\mathbb{R}^\mathbb{N}\ = \{(a_0, a_1, a_2, \dots)|a_i \in \mathbb{R} \text{ for all }i \in \mathbb{N}\}\] and the set of bounded functions on the real line \[B(\mathbb{R}) = \{f:\mathbb{R} \rightarrow \mathbb{R} | \operatorname{sup}_{x \in \mathbb{R}}{|f(x)|} < \infty\}\] as real vector spaces cannot be spanned by a countable subset. This is a consequence of the Baire Category Theorem in functional analysis. ∎
To work with infinite dimensional spaces, it is more convenient to characterize bases in a different way. A linearly independent subset is called maximal iff it is not a proper subset of any linearly independent subset.
Proposition. A subset is a basis iff it is a maximal linearly independent subset.
Proof. Let $S$ be a linearly independent subset of a vector space $V$. $S$ does not span $V$ iff there exists $v\in V-\operatorname{span}(S)$ iff there exists $v\in V$ such that $S\cup \{v\}$ is linearly independent iff $S$ is not maximal. ∎
We can find a maximal linearly independent subset in any vector space using the Zorn's lemma, which is logically equivalent to the axiom of choice in set theory.
Zorn's Lemma. If $\mathcal{S}$ is a nonempty partially ordered set in which every chain (totally ordered subset) has an upper bound in $\mathcal{S}$, then $\mathcal{S}$ has at least one maximal element. ∎
Zorn's Lemma. If $\mathcal{S}$ is a nonempty partially ordered set in which every chain (totally ordered subset) has an upper bound in $\mathcal{S}$, then $\mathcal{S}$ has at least one maximal element. ∎
Theorem. Every vector space has a basis.
Proof. Let $\mathcal{S}$ be the set of linearly independent subsets in the vector space, partially ordered by inclusion. ($\varnothing \in \mathcal{S}$ so $\mathcal{S}$ is nonempty.) We will show, by Zorn's Lemma, that $\mathcal{S}$ has a maximal element, i.e. a maximal linearly independent subset. Then this subset is a basis for the vector space.
We verify the hypothesis of Zorn's Lemma. Given a chain $C = \{S_\alpha|\alpha \in I\}$ in $\mathcal{S}$, we show that $S_I = \cup_{\alpha\in I}{S_\alpha}$ is linearly independent, hence an upper bound of $C$. For any finite subset $\{v_1, v_2, \dots, v_n\}$ of $S_I$, there are $\alpha_1, \alpha_2, \dots, \alpha_n \in I$ such that $v_i \in S_{\alpha_i}$. Since $\{S_{\alpha_i}\}$ is totally ordered, there is a greatest element $S_\alpha$ among them, i.e. $S_{\alpha_i}\subseteq S_\alpha$ for all $i$. Then $\{v_1, v_2, \dots, v_n\}\subseteq S_\alpha$ is a linearly independent subset. By definition, $S_I$ is linearly independent. ∎
The next proposition is a slight modification of the theorem.
Proposition.
(a) Every linearly independent subset can be extended to a basis.
(b) Every spanning subset contains a basis.
Proof. Let $S$ be a spanning subset and $L$ be a linearly independent subset.
(a) In the previous proof, replace $\mathcal{S}$ by the set of linearly independent subsets containing $L$.
(b) Replace $\mathcal{S}$ by the set of linearly independent subsets contained in $S$ and obtain a maximal element $M$ (which is maximal in the sense that it is not a proper subset of another linearly independent subset contained in $S$). Assume that $M$ is not a maximal linearly independent subset, then there exists $v \in V-\operatorname{span}(M)$. Write $v=c_1 v_1 + c_2 v_2 + \dots + c_n v_n$ where $\{v_1, v_2, \dots, v_n\}\subseteq S$. Some $v_i \not\in \operatorname{span}(M)$, say $v_1$, then $M \cup \{v_1\}$ is a linearly independent subset strictly containing $M$, contradicts the maximality of $M$. ∎
A result in cardinal arithmetic is needed to proceed.
(Cardinals (or cardinal numbers) are a measure of the cardinality of sets. A cardinal is a natural choice of set that represent all sets with the same cardinality as it. To be more specific, $\operatorname{card}(A)$ is a cardinal that has the same cardinality as the set $A$.)
Proposition. If $\kappa$ is an infinite cardinal, then $\kappa\cdot\kappa=\kappa$, where $\kappa\cdot\kappa$ is defined to be $\operatorname{card}(\kappa\times\kappa)$. ∎
In other words, if a set $A$ is infinite, then $\operatorname{card}(A \times A) = \operatorname{card}(A)$. The fact that $\mathbb{N} \times \mathbb{N}$ is countable is included as a special case.
Remark. Writing the Cartesian product as a union, it follows that the union of at most $\kappa$ many sets, each of cardinality at most $\kappa$, has cardinality at most $\kappa$. Also, apply the proposition several times we get $\kappa^n = \kappa$ for all $n \in \mathbb{N}-\{0\}$.∎
Here comes the main fact:
Theorem. The cardinality of a spanning subset has at least the cardinality of a linearly independent subset.
Proof. Let $S$ be a spanning subset and $L$ be a linearly independent subset of a vector space $V$. What we need to show is $\operatorname{card}(S)\geq\operatorname{card}(L)$.
Extend $L$ to a basis $L'$, then $\operatorname{card}(L) \leq \operatorname{card}(L')$. Every $v \in S$ can be expressed as a linear combination of some finite subset $L'_v$ of $L'$, so $V = \operatorname{span}(S) \subseteq \operatorname{span}(\cup_{v \in S}{L'_v})$. As $L'$ is a maximal linearly independent subset, no proper subset of it can span $V$, hence $\cup_{v \in S}{L'_v} = L'$. By the remark, $\operatorname{card}(L) \leq \operatorname{card}(L') = \operatorname{card}(\cup_{v \in S}{L'_v}) \leq \operatorname{card}(S)$. ∎
Corollary. Any two bases for a vector space has the same cardinality.
Proof. Apply the previous theorem twice and the Cantor-Bernstein-Schröder Theorem in set theory. ∎
Finally, we include two more results. The first generalizes the technique used in example (b), which relates the cardinality and the dimension of a vector space. The second constructs vector spaces of arbitrary dimension and describes all vector spaces.
Proposition. Let $V$ be a vector space over a field $F$ and $\operatorname{dim}V$ be its dimension over $F$.
(a) If $V$ is finite dimensional, then $\operatorname{card}(V) = \operatorname{card}(F)^{\operatorname{dim}V}$. In particular, if $F$ is infinite, then $\operatorname{card}(V) = \operatorname{card}(F)$.
(b) If $V$ is infinite dimensional, then $\operatorname{card}(V) = \max{\{\operatorname{card}(F), \operatorname{dim}V\}}$.
Proof.
(a) The first part is just typical linear algebra, and the second part follows from $\operatorname{card}(F^{\operatorname{dim}V}) = \operatorname{card}(F)^{\operatorname{dim}V} = \operatorname{card}(F)$ by the remark.
(b) Let $B$ be a basis for $V$. Then $V$ is in one-one correspondence with the set of finite subsets of $(F - \{0\}) \times B$. By the remark, since $(F-\{0\}) \times B$ is infinite (as $B$ is), the latter set has cardinality $\operatorname{card}((F-\{0\}) \times B)$, which equals $\max{\{\operatorname{card}(F-\{0\}), \operatorname{card}(B)\}} = \max{\{\operatorname{card}(F), \operatorname{card}(B)\}}$$= \max{\{\operatorname{card}(F), \operatorname{dim}V\}}$ by the previous proposition. ∎
Definition. Given a field $F$ and an arbitrary set $X$. The set of functions from $X$ to $F$ which vanishes outside a finite subset of $X$, \[(F^X)_0 = \{f: X \rightarrow F | f^{-1}(F-\{0\}) \text{ is finite}\},\]is a vector space over $F$ and is called a coordinate space. ∎
The polynomial ring in one variable is isomorphic to $(F^\mathbb{N})_0$ as vector space.
Proposition. Given a field $F$.
(a) The coordinate space $(F^X)_0$ has dimension $\operatorname{card}(X)$ over $F$. Hence there exist vector spaces over $F$ of arbitrary dimension.
(b) Every vector space over $F$ is isomorphic to a coordinate space over $F$.
Proof.
(a) $(F^X)_0$ has a canonical basis $\{f_x\}_{x \in X}$ where $f_x(x) = 1$ and $f_x(y) = 0$ for all $y \in X -\{x\}$. Given an arbitrary cardinal $\kappa$, the vector space $(F^\kappa)_0$ has dimension $\kappa$.
(b) Let $V$ be a vector space over $F$ and $B$ be a basis for $V$. The linear map $\Phi: V \rightarrow (F^B)_0$ which sends $v \in B$ to $f_v \in (F^B)_0$ is an isomorphism. ∎
