Bases and Dimension: Examples and Counterexamples

Coordinates in $e$. By the standard basis lemma, $e$ is an ordered basis of $\mathbb{R}^{2}$ and the coordinate list of $y \in \mathbb{R}^{2}$ is $i \mapsto y(i)$; for $x = (1,2)$ that list is $(1,2)$.

$e'$ is an ordered basis. The list $e' = (e_1, e_0)$ is injective, since $e_0 \ne e_1$ ($e_0$ takes the value $1$ at $0$ and $e_1$ takes the value $0$ there), and its image is $\{e_0, e_1\} = e[2]$, which is a basis of $\mathbb{R}^{2}$; so $e'$ is an injective list whose image is a basis.

Coordinates in $e'$. For $\lambda : 2 \to \mathbb{R}$, $\sum_{i<2}\lambda_i e'_i = \lambda_0 e_1 + \lambda_1 e_0 = \lambda_1 e_0 + \lambda_0 e_1 = \sum_{i<2}\mu_i e_i$ with $\mu_0 = \lambda_1$ and $\mu_1 = \lambda_0$; evaluating with the standard basis, this vector is $(\lambda_1, \lambda_0)$. It equals $x = (1,2)$ exactly when $\lambda_1 = 1$ and $\lambda_0 = 2$, so the coordinate list of $x$ in $e'$ is $(2,1)$.

$v$ is an ordered basis and the coordinates of a general vector in it. Note $v_0 = (1,1) = e_0 + e_1$ and $v_1 = (1,0) = e_0$. For $\lambda : 2 \to \mathbb{R}$, $\sum_{i<2}\lambda_i v_i = \lambda_0(e_0+e_1) + \lambda_1 e_0 = (\lambda_0 e_0 + \lambda_0 e_1) + \lambda_1 e_0 = (\lambda_0 + \lambda_1)e_0 + \lambda_0 e_1$, using (V2), (V3) and the abelian group laws; by the standard basis this vector is $(\lambda_0 + \lambda_1,\ \lambda_0)$. Given $y = (y_0, y_1)$, the equations $\lambda_0 + \lambda_1 = y_0$ and $\lambda_0 = y_1$ have the unique solution $\lambda_0 = y_1$, $\lambda_1 = y_0 - y_1$, so every $y$ is $\sum_{i<2}\lambda_i v_i$ for exactly one $\lambda$ and $v$ is an ordered basis.

Coordinates of $x$ in $v$. Taking $y = x = (1,2)$ in step 1.4 gives $\lambda_0 = 2$ and $\lambda_1 = 1 - 2 = -1$, so the coordinate list of $x$ in $v$ is $(2,-1)$; and $2(1,1) + (-1)(1,0) = (2,2) + (-1,0) = (1,2) = x$ confirms it.

The three coordinate lists of the single vector $x$ are therefore $(1,2)$, $(2,1)$ and $(2,-1)$, computed in steps 1.1, 1.3 and 2.1; the first two are different although $e$ and $e'$ have the same image, so the coordinate list depends on the ordered basis and not merely on the underlying set.

Claim 1. $E$ is nonempty: the zero family has value $0_F$ everywhere, so $N = 0$ witnesses that it lies in $E$. And $E$ is closed under the one-step expression: for $\lambda \in F$ and $u, v \in E$ with witnesses $N_1$ and $N_2$, let $N$ be the larger of the two, which exists because the order of $\mathbb{N}$ is total; then for $n \ge N$ we have $n \ge N_1$ and $n \ge N_2$, so $(\lambda u + v)(n) = \lambda u(n) + v(n) = \lambda 0_F + 0_F = 0_F$, and $N$ witnesses $\lambda u + v \in E$. So $E$ is a linear subspace of $F^{\mathbb{N}}$ by the one-step test.

Each $e_k$ lies in $E$, so $B \subseteq E$: if $n \ge \sigma(k)$ then $k < n$, hence $n \ne k$ and $e_k(n) = 0_F$, so $\sigma(k)$ is a witness.

For $x \in F^{\mathbb{N}}$ and $N \in \mathbb{N}$ put $s_N := \sum_{k<N} x(k)\,e_k$. Then $s_N(n) = x(n)$ for $n < N$ and $s_N(n) = 0_F$ for $n \ge N$. Indeed $s_N(n) = \sum_{k<N}\bigl(x(k)e_k\bigr)(n) = \sum_{k<N} x(k)\,e_k(n)$ by pointwise evaluation and pointwise scalar multiplication; the scalar list $k \mapsto x(k)e_k(n)$ has the value $x(k)0_F = 0_F$ at every $k \ne n$. If $n < N$ this list vanishes off the single index $n$, where its value is $x(n)1_F = x(n)$, so the sum is $x(n)$; if $n \ge N$ then no $k < N$ equals $n$, the list is all $0_F$, and the sum is $0_F$.

Claim 3, independence. The map $k \mapsto e_k$ is injective, since $e_k(k) = 1_F \ne 0_F = e_l(k)$ for $l \ne k$. Let $w : m \to B$ be an injective finite list and $\lambda : m \to F$ with $\sum_{j<m}\lambda_j w_j = 0$ in $F^{\mathbb{N}}$. Each $w_j$ is $e_{i_j}$ for exactly one $i_j \in \mathbb{N}$, and $j \mapsto i_j$ is injective because $w$ is. Fix $j < m$ and evaluate at $i_j$: pointwise evaluation gives $0_F = \sum_{l<m}\lambda_l\,e_{i_l}(i_j)$, and $e_{i_l}(i_j) = 0_F$ unless $i_l = i_j$, that is unless $l = j$, where it is $1_F$. So the scalar list vanishes off the single index $j$ and sums to $\lambda_j 1_F = \lambda_j$, giving $\lambda_j = 0_F$. Hence every injective finite list into $B$ is independent, that is $B$ is linearly independent.

$B \approx \mathbb{N}$: the map $k \mapsto e_k$ is injective by step 1.4 and its image is $B$ by definition, so it is a bijection $\mathbb{N} \to B$.

Claim 2. Each $e_k$ lies in $E$ and $E$ is a linear subspace, so $\operatorname{span}(B) \subseteq E$ by minimality of the span. Conversely let $x \in E$ with witness $N$; then $x$ and $s_N$ of step 1.3 agree at every $n$, since for $n < N$ both take the value $x(n)$ and for $n \ge N$ both take the value $0_F$, so $x = s_N = \sum_{k<N} x(k)e_k$, a linear combination of elements of $B$. Hence $E \subseteq \operatorname{span}(B)$.

Claim 4. Suppose $E$ had a finite basis $C$, say with $p$ elements. Then $C$ is a spanning set of $E$ with $p$ elements, so no linearly independent subset of $E$ is equinumerous with $\mathbb{N}$. But $B \subseteq E$ is linearly independent by step 1.4 and $B \approx \mathbb{N}$ by step 2.1. So no finite basis exists and $E$ is infinite-dimensional over $F$.

Claim 3, that $B$ is a basis of $E$. By step 1.4 the set $B$ is linearly independent and by step 2.2 it spans $E$; independence and spans computed in the linear subspace $E$ agree with those computed in $F^{\mathbb{N}}$, so $B$ is a basis of the vector space $E$.

Claim 1. $\mathbb{Q}_{\mathbb{R}} = \iota[\mathbb{Q}]$ is a subfield of $\mathbb{R}$: it contains $\iota(1) = 1$; for $p, q \in \mathbb{Q}$ it contains $\iota(p) - \iota(q) = \iota(p-q)$ and $\iota(p)\iota(q) = \iota(pq)$; and if $\iota(q) \ne 0$ then $q \ne 0$, since $\iota(0) = 0$, so $\iota(q)^{-1} = \iota(q^{-1}) \in \mathbb{Q}_{\mathbb{R}}$. Since $\mathbb{R}$ is a vector space over itself, restriction of scalars makes it a vector space over $\mathbb{Q}_{\mathbb{R}}$, with the field multiplication restricted to $\mathbb{Q}_{\mathbb{R}} \times \mathbb{R}$. The operation $(q,x) \mapsto \iota(q)x$ is a map $\mathbb{Q} \times \mathbb{R} \to \mathbb{R}$, and it satisfies (V2) to (V5) because $\iota$ preserves sums and products and $\iota(1) = 1$, while (V1) is the abelian group $(\mathbb{R},+,0)$; so it makes $\mathbb{R}$ a vector space over $\mathbb{Q}$.

$\mathbb{Q}_{\mathbb{R}}$ is at most countable: $\iota$ is injective with image $\mathbb{Q}_{\mathbb{R}}$, hence a bijection $\mathbb{Q} \to \mathbb{Q}_{\mathbb{R}}$, and $\mathbb{Q} \approx \mathbb{N}$; composing bijections gives $\mathbb{Q}_{\mathbb{R}} \approx \mathbb{N}$.

If $K$ is at most countable then so is $K^{n}$, the set of functions $n \to K$, for every $n \in \mathbb{N}$. By induction on $n$: at $n = 0$ the set $K^{0}$ has exactly one element, the empty function, so it is finite; and the map $K^{\sigma(n)} \to K^{n} \times K$ sending $f$ to the pair consisting of its restriction to $n$ and its value at $n$ is a bijection, since $\sigma(n) = n \cup \{n\}$ and $n \notin n$, so a function on $\sigma(n)$ is determined by, and may be assembled from, those two data. Hence $K^{\sigma(n)} \approx K^{n} \times K$, which is at most countable by the inductive hypothesis and the product theorem, and countability transfers along the bijection.

Claim 4. For a list $v : n \to \mathbb{R}$ and scalars $\lambda : n \to \mathbb{Q}$, the vector $\sum_{i<n}\lambda_i \cdot v_i$ computed in the $\mathbb{Q}$-structure is by definition $\sum_{i<n}\iota(\lambda_i)v_i$, computed in the $\mathbb{Q}_{\mathbb{R}}$-structure; the two structures have the same underlying set, the same addition and the same zero, so their finite sums agree. Since $\iota$ is a bijection $\mathbb{Q} \to \mathbb{Q}_{\mathbb{R}}$, the scalar lists $\lambda : n \to \mathbb{Q}$ and $\iota \circ \lambda : n \to \mathbb{Q}_{\mathbb{R}}$ correspond bijectively, and $\lambda_i = 0$ for all $i$ exactly when $\iota(\lambda_i) = 0$ for all $i$. So a vanishing combination exists on one side exactly when it does on the other, and likewise for representations of an arbitrary vector; hence the two structures have the same linearly independent subsets, the same spans and the same bases.

Claim 2. $\mathbb{R}$ is a vector space over $\mathbb{Q}_{\mathbb{R}}$ by step 1.1, and every vector space has a basis, so a basis $B$ of $\mathbb{R}$ over $\mathbb{Q}_{\mathbb{R}}$ exists.

Claim 3. Suppose some basis $B$ of $\mathbb{R}$ over $\mathbb{Q}_{\mathbb{R}}$ were finite, say $B \approx n$. A bijection $n \to B$ is an injective list whose image is a basis, hence an ordered basis, so every $x \in \mathbb{R}$ is $\sum_{i<n}\lambda_i b_i$ for exactly one $\lambda : n \to \mathbb{Q}_{\mathbb{R}}$. The resulting map $\Phi : \mathbb{R} \to (\mathbb{Q}_{\mathbb{R}})^{n}$, sending $x$ to that $\lambda$, is injective, since $\Phi(x) = \Phi(y)$ makes $x$ and $y$ the same sum. By steps 1.2 and 1.3 the set $(\mathbb{Q}_{\mathbb{R}})^{n}$ is at most countable, hence so is its subset $\Phi[\mathbb{R}]$; and $\Phi$ is a bijection $\mathbb{R} \to \Phi[\mathbb{R}]$, so $\mathbb{R}$ is at most countable, contradicting the uncountability of $\mathbb{R}$. So no basis of $\mathbb{R}$ over $\mathbb{Q}_{\mathbb{R}}$ is finite, and $\mathbb{R}$ is infinite-dimensional over $\mathbb{Q}_{\mathbb{R}}$.

Claim 1 is step 1.1, claim 2 is step 2.1, claim 3 is step 2.2, and claim 4 is step 1.4; by claim 4 the last two transfer to $\mathbb{R}$ as a $\mathbb{Q}$-vector space.