Proof of The Span of a Finite Family is the Smallest Subspace Containing It

Sums of vectors are finite sums in $V$, and we use Properties of Finite Sums of Vectors together with the identities of Elementary Identities in a Vector Space, in particular $0\,x=0_{V}$ and $1\,x=x$ for $x\in V$, where $0$ and $1$ are the additive and multiplicative identities of $K$. By the definition of a tuple, an $n$-tuple in a set $X$ is a map from $[n]$ to $X$, so the finite sums below are formed from maps on $[n]$ as required. Write $Z=\operatorname{span}(v)$.

Claim 1. $0_{V}\in Z$. Let $c\in K^{n}$ have $c_{k}=0$ for every $k\in[n]$. Then $c_{k}v_{k}=0_{V}$ for every $k$, so $\sum_{k=1}^{n}c_{k}v_{k}=0_{V}$ by claim 7 of Properties of Finite Sums of Vectors; hence $0_{V}\in Z$.

Closure under addition. Let $u=\sum_{k=1}^{n}c_{k}v_{k}$ and $u'=\sum_{k=1}^{n}c'_{k}v_{k}$ lie in $Z$, with $c,c'\in K^{n}$. By the distributivity axiom of Vector Space over a Field we have $(c_{k}+c'_{k})v_{k}=c_{k}v_{k}+c'_{k}v_{k}$ for every $k$, so claim 2 of Properties of Finite Sums of Vectors gives

and the components $c_{k}+c'_{k}$ form an $n$-tuple in $K$, so $u+u'\in Z$.

Closure under scalar multiplication. Let $\lambda\in K$ and $u=\sum_{k=1}^{n}c_{k}v_{k}\in Z$ with $c\in K^{n}$. By claim 3 of Properties of Finite Sums of Vectors and the axiom $\lambda(\mu x)=(\lambda\mu)x$ of Vector Space over a Field,

and the components $\lambda c_{k}$ form an $n$-tuple in $K$, so $\lambda u\in Z$. Thus $Z$ satisfies the three conditions of Linear Subspace and is a linear subspace of $V$.

The components of $v$ lie in $Z$. Fix $j\in[n]$ and let $c\in K^{n}$ have $c_{j}=1$ and $c_{k}=0$ for $k\ne j$. Then $c_{k}v_{k}=0_{V}$ for every $k\ne j$ and $c_{j}v_{j}=v_{j}$, so $\sum_{k=1}^{n}c_{k}v_{k}=v_{j}$ by claim 7 of Properties of Finite Sums of Vectors, and $v_{j}\in Z$.

Claim 2. By claim 1 and claim 1 of A Linear Subspace is a Vector Space and Inherits an Inner Product, $Z$ is a vector space over $K$ under the restricted operations, and by claim 2 of that lemma a finite sum of an $n$-tuple with components in $Z$ has the same value whether formed in $Z$ or in $V$. Let $u\in Z$. By Span of a Finite Family of Vectors there is $c\in K^{n}$ with $u=\sum_{k=1}^{n}c_{k}v_{k}$, the sum formed in $V$. Each $c_{k}v_{k}$ lies in $Z$ by claim 1, and the scalar multiplication of $Z$ is that of $V$, so the same equation reads as an identity of finite sums formed in $Z$ with the $n$-tuple $v^{\ast}$. Hence every element of $Z$ is such a combination, which is what it means for $v^{\ast}$ to span $Z$.

Claim 3. Let $W$ be a linear subspace of $V$ with $v_{k}\in W$ for every $k\in[n]$, and let $u\in Z$, say $u=\sum_{k=1}^{n}c_{k}v_{k}$ with $c\in K^{n}$. For a natural number $i$ let $P(i)$ be the assertion: if $i\in[n]$, then $\sum_{k=1}^{i}c_{k}v_{k}\in W$. We prove $P(i)$ for every $i$ by induction.

For $i=1$: if $1\in[n]$, then $\sum_{k=1}^{1}c_{k}v_{k}=c_{1}v_{1}$ by claim 1 of Properties of Finite Sums of Vectors, and this lies in $W$ because $v_{1}\in W$ and $W$ is closed under scalar multiplication.

Assume $P(i)$ and suppose $S(i)\in[n]$, where $S$ is the successor map of Natural Numbers. Then $i<S(i)$ and $S(i)\le n$ give $i\le n$, and $1\le i$, so $i\in[n]$ by Properties of the Order on the Natural Numbers. By the recursion in claim 1 of Properties of Finite Sums of Vectors,

The first summand lies in $W$ by $P(i)$ and the second lies in $W$ because $v_{S(i)}\in W$ and $W$ is closed under scalar multiplication; hence the sum lies in $W$ because $W$ is closed under addition. This proves $P(S(i))$.

Taking $i=n$ gives $u\in W$, so $Z\subseteq W$.