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 $(1,2)$.

$e^{\mathrm{\prime }}$ is an ordered basis. The list $e^{\mathrm{\prime }}=(e_1,e_0)$ is injective, since $e_0\mathrm{\ne }{e}_1$ ($e_0$ 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 $e^{\mathrm{\prime }}$ is an injective list whose image is a basis.

Coordinates in $e^{\mathrm{\prime }}$. For $\lambda :2\to \mathbb{R}$, ${\sum }_{i<2}{\lambda }_i{e}_i^{\mathrm{\prime }}={\lambda }_0{e}_1+{\lambda }_1{e}_0={\lambda }_1{e}_0+{\lambda }_0{e}_1={\sum }_{i<2}{\mu }_i{e}_i$ 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 $e^{\mathrm{\prime }}$ 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 $e^{\mathrm{\prime }}$ 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 $n\mathrm{\ne }k$ 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 $s_N(n)={\sum }_{k<N}(x(k)e_k)(n)={\sum }_{k<N}x(k)e_k(n)$ 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 $k\mathrm{\ne }n$. 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 $e_k(k)=1_F\mathrm{\ne }{0}_F=e_l(k)$ for $l\mathrm{\ne }k$. 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 $\mathrm{span}(B)\subseteq E$ 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 $E\subseteq \mathrm{span}(B)$.

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 $\iota (q)\mathrm{\ne }0$ then $q\mathrm{\ne }0$, 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 $n\mathrm{\notin }n$, 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 $\mathrm{\Phi }:\mathbb{R}\to ({\mathbb{Q}}_{\mathbb{R}}{)}^n$, sending $x$ to that $\lambda$, is injective, since $\mathrm{\Phi }(x)=\mathrm{\Phi }(y)$ 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 $\mathrm{\Phi }[\mathbb{R}]$; and $\mathrm{\Phi }$ is a bijection $\mathbb{R}\to \mathrm{\Phi }[\mathbb{R}]$, 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.

$B$ is linearly independent, $\mathrm{span}(B)=E$, and $B\approx \mathbb{N}$.

$B$ is not finite: if $B\approx p$ for some $p\in \mathbb{N}$ then, since $B\approx \mathbb{N}$, symmetry and transitivity of $\approx$ would give $\mathbb{N}\approx p$, which is impossible.

The constant family $c$ with $c(n)=1_F$ for every $n$ lies in $F^{\mathbb{N}}$ and not in $E$: for any candidate witness $N$ we have $N\ge N$ and $c(N)=1_F\mathrm{\ne }{0}_F$, so no $N$ witnesses that $c$ is eventually zero.

$F^{\mathbb{N}}$ is infinite-dimensional. If it had a finite basis, that basis would be a spanning set with $p$ elements for some $p$, and then no linearly independent subset of $F^{\mathbb{N}}$ would be equinumerous with $\mathbb{N}$; but $B$ is such a subset.

So $B$ is a linearly independent subset of the infinite-dimensional space $F^{\mathbb{N}}$, it is not finite, and $\mathrm{span}(B)=E\mathrm{\ne }{F}^{\mathbb{N}}$, since $c$ lies in the second and not the first. The false claim therefore fails, and $B$ is not a basis of $F^{\mathbb{N}}$, its span not being the whole space.