Vector Spaces and Linear Subspaces: Examples and Counterexamples

$\mathbb{R}$ is a field, so it is a vector space over itself with the field addition as vector addition and the field multiplication as scalar multiplication; this is claim 1.

Since $\mathbb{R}$ is an ordered field, there is a unique field homomorphism $\iota : \mathbb{Q} \to \mathbb{R}$, and it is injective.

$\mathbb{Q}_{\mathbb{R}} = \iota[\mathbb{Q}]$ is a subfield of $\mathbb{R}$: it contains $\iota(0) = 0$ and $\iota(1) = 1$; for $p, q \in \mathbb{Q}$ it contains $\iota(p) + \iota(q) = \iota(p+q)$, $-\iota(q) = \iota(-q)$ and $\iota(p)\iota(q) = \iota(pq)$; and if $\iota(q) \ne 0$ then $q \ne 0$, since $\iota(0) = 0$, so $\iota(q)^{-1} = \iota(q^{-1}) \in \mathbb{Q}_{\mathbb{R}}$.

The assignment $(q,x) \mapsto \iota(q)\,x$ is a map $\mathbb{Q} \times \mathbb{R} \to \mathbb{R}$, since $\iota(q) \in \mathbb{R}$ and the field multiplication of $\mathbb{R}$ takes values in $\mathbb{R}$.

Applying restriction of scalars to the $\mathbb{R}$-vector space of step 1.1 and the subfield of step 1.3 shows that $\mathbb{R}$ is a vector space over $\mathbb{Q}_{\mathbb{R}}$, with the field multiplication restricted to $\mathbb{Q}_{\mathbb{R}} \times \mathbb{R}$ as scalar multiplication; this is claim 2.

The operation of step 1.4 satisfies the five axioms over $\mathbb{Q}$. (V1) holds because $(\mathbb{R},+,0)$ is an abelian group. For $p, q \in \mathbb{Q}$ and $x, y \in \mathbb{R}$: $q \cdot (x+y) = \iota(q)(x+y) = \iota(q)x + \iota(q)y = q\cdot x + q\cdot y$ by distributivity, which is (V2); $(p+q)\cdot x = \iota(p+q)x = (\iota(p)+\iota(q))x = \iota(p)x + \iota(q)x = p\cdot x + q\cdot x$ by additivity of $\iota$ and distributivity, which is (V3); $(pq)\cdot x = \iota(pq)x = (\iota(p)\iota(q))x = \iota(p)(\iota(q)x) = p\cdot(q\cdot x)$ by multiplicativity of $\iota$ and associativity, which is (V4); and $1 \cdot x = \iota(1)x = 1\,x = x$, which is (V5).

Claim 1 is step 1.1, claim 2 is step 2.1, and claim 3 is step 2.2, so $\mathbb{R}$ carries all three structures at once: over itself, over the embedded copy of $\mathbb{Q}$ inside it, and over $\mathbb{Q}$.

$F^{3}$ is the set of functions $3 \to F$ with the pointwise operations, and $3 = \{0,1,2\}$, so the coordinates of an element are $x_0, x_1, x_2$.

Each $e_j$ is an element of $F^{3}$, and each $L_j$ is a subset of $F^{3}$, both by their displayed descriptions.

$L_j = \{\, \lambda e_j : \lambda \in F \,\}$ for every $j < 3$. If $\lambda \in F$ then $(\lambda e_j)_j = \lambda 1_F = \lambda$ and $(\lambda e_j)_i = \lambda 0_F = 0_F$ for $i \ne j$, so $\lambda e_j \in L_j$. Conversely if $x \in L_j$ then $x$ and $x_j e_j$ have the same value at every $i < 3$, namely $x_j$ at $i = j$ and $0_F$ elsewhere, so $x = x_j e_j$.

For $u_0, u_1, u_2 \in F^{3}$ the finite sum is $\sum_{j<3} u_j = (u_0 + u_1) + u_2$, whose value at $i < 3$ is $(u_0(i) + u_1(i)) + u_2(i)$, by the pointwise definition of the addition.

$F^{0}$ has exactly one element, the empty function, and that element is its zero vector, so $F^{0}$ is the zero space; this is claim 3.

Each $L_j$ is a linear subspace of $F^{3}$ and equals $\operatorname{span}\{e_j\}$: by step 1.3 it is the set of scalar multiples of $e_j$, which is exactly the span of $\{e_j\}$, and a span is a linear subspace. This is claim 1.

Every $x \in F^{3}$ decomposes. Put $u_j := x_j e_j$, which lies in $L_j$ by step 1.3. The value of $\sum_{j<3} u_j$ at $i < 3$ is $(x_0 e_0(i) + x_1 e_1(i)) + x_2 e_2(i)$; since $e_j(i) = 0_F$ for $j \ne i$ and $e_i(i) = 1_F$, exactly one summand is $x_i$ and the others are $0_F$, so the value is $x_i$. Hence $\sum_{j<3} u_j = x$, and $\sum_{j<3} L_j = F^{3}$.

The decomposition is unique. Suppose $u_j \in L_j$ for $j < 3$ and $\sum_{j<3} u_j = x$. Evaluating at $i < 3$ gives $(u_0(i) + u_1(i)) + u_2(i) = x_i$, and $u_j(i) = 0_F$ whenever $j \ne i$, so the left-hand side is $u_i(i)$; thus $u_i(i) = x_i$, and step 1.3 gives $u_i = u_i(i) e_i = x_i e_i$. So the list is the one of step 2.2.

By steps 2.2 and 3.1 every $x \in F^{3}$ is $\sum_{j<3} u_j$ with $u_j \in L_j$ in exactly one way, so $F^{3} = \bigoplus_{j<3} L_j$, and the decomposition is $x = x_0 e_0 + x_1 e_1 + x_2 e_2$. This is claim 2.

Claim 1 is step 2.1, claim 2 is step 4.1 and claim 3 is step 1.5.

$F^{\mathbb{N}}$ is a vector space over $F$, being the function space on the index set $\mathbb{N}$, and its zero is the constant family at $0_F$.

$E$ is nonempty: the zero family has $x_n = 0_F$ for every $n$, so $N = 0$ witnesses that it lies in $E$.

$E$ is closed under the one-step expression. Let $\lambda \in F$ and $u, v \in E$, with witnesses $N_1$ for $u$ and $N_2$ for $v$. The order of $\mathbb{N}$ is total, so one of $N_1, N_2$ is at least the other; let $N$ be that one. For $n \ge N$ we have $n \ge N_1$ and $n \ge N_2$, hence $(\lambda u + v)_n = \lambda u_n + v_n = \lambda 0_F + 0_F = 0_F$. So $N$ witnesses $\lambda u + v \in E$.

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

For $x \in F^{\mathbb{N}}$ and $N \in \mathbb{N}$, the finite sum $s_N := \sum_{k<N} x_k e_k$ satisfies $s_N(n) = x_n$ for $n < N$ and $s_N(n) = 0_F$ for $N \le n$. By induction on $N$: at $N = 0$ the sum is the zero family, there is no $n < 0$, and the second clause holds. Assuming it at $N$, we have $s_{\sigma(N)} = s_N + x_N e_N$, so $s_{\sigma(N)}(n) = s_N(n) + x_N e_N(n)$; for $n < N$ this is $x_n + x_N 0_F = x_n$, since $n \ne N$; for $n = N$ it is $0_F + x_N 1_F = x_N$; and for $\sigma(N) \le n$ we have $N \le n$ and $n \ne N$, so it is $0_F + x_N 0_F = 0_F$. As $n < \sigma(N)$ is equivalent to $n \le N$, this is the claim at $\sigma(N)$.

The constant family $c$ with $c_n = 1_F$ for every $n$ does not lie in $E$: for any candidate witness $N$ we have $N \ge N$ and $c_N = 1_F \ne 0_F$. Hence $E \ne F^{\mathbb{N}}$, which is claim 3.

$E$ is a linear subspace of $F^{\mathbb{N}}$, by the one-step test applied to steps 1.2 and 1.3; this is claim 1.

If $x \in E$ with witness $N$, then $x = \sum_{k<N} x_k e_k$. Indeed the two families agree at every $n$: at $n < N$ both are $x_n$, and at $N \le n$ the sum is $0_F$ while $x_n = 0_F$ by the choice of $N$. So $x$ is a linear combination of elements of $\{\, e_k : k \in \mathbb{N} \,\}$.

Claim 2. By step 2.2 every element of $E$ is a linear combination of elements of $\{\, e_k : k \in \mathbb{N} \,\}$, hence lies in its span. Conversely each $e_k$ lies in $E$ by step 1.4 and $E$ is a linear subspace by step 2.1, so the span, being the smallest linear subspace containing all the $e_k$, is contained in $E$.

Claim 1 is step 2.1, claim 2 is step 3.1 and claim 3 is step 1.6.

$\mathbb{R}^{2}$ is the vector space of functions $2 \to \mathbb{R}$ with coordinatewise operations, $2 = \{0,1\}$, so an element is $x = (x_0, x_1)$ and the zero vector is $(0,0)$.

The zero vector lies in $Q$, since $0 \le 0$; in particular $Q$ is nonempty.

$0 < 1$ in $\mathbb{R}$: $1 \ne 0$ and $1 = 1 \cdot 1$ is a square, so $1 \in P$.

If $a, b \in \mathbb{R}$ satisfy $0 \le a$ and $0 \le b$, then $0 \le a + b$: if $a = 0$ then $a + b = b$; if $b = 0$ then $a + b = a$; and otherwise $a, b \in P$, so $a + b \in P$ by (O2).

$Q$ is closed under addition: for $x, y \in Q$ and $i \in \{0,1\}$ we have $(x+y)_i = x_i + y_i$ with $0 \le x_i$ and $0 \le y_i$, hence $0 \le (x+y)_i$.

It is not the case that $0 \le -1$: applying trichotomy to $-1$, exactly one of $-1 \in P$, $-1 = 0$, $-(-1) = 1 \in P$ holds, and the last one does, so $-1 \notin P$ and $-1 \ne 0$.

The vector $e_0$ with $(e_0)_0 = 1$ and $(e_0)_1 = 0$ lies in $Q$, since $0 \le 1$ and $0 \le 0$.

$(-1)e_0$ has coordinates $((-1)\cdot 1, (-1)\cdot 0) = (-1, 0)$, and its coordinate at index $0$ fails $0 \le -1$, so $(-1)e_0 \notin Q$. Hence $Q$ is not closed under scalar multiplication: condition (W3) fails, and $Q$ is not a linear subspace of $\mathbb{R}^{2}$.

So $Q$ contains the zero vector and is closed under addition, by steps 1.2 and 2.1, and is not a linear subspace, by step 3.1; the claim of [L8] is therefore false. The failure can also be read in the additive group: $(-1)e_0 = -e_0$ lies outside $Q$, so $Q$ is not even a subgroup of the additive group of $\mathbb{R}^{2}$, whereas a linear subspace always is.