Proof of Extraction of a Summand from a Finite Sum of Vectors

All sums are finite sums in $V$. We use claim 1 of Properties of Finite Sums of Vectors, which gives $\sum_{k=1}^{1}x_{k}=x_{1}$, the recursion

(the successor of a natural number $p$ being $p+1$ by Natural Numbers), and the fact that a finite sum is unchanged when the tuple of summands is replaced by a restriction of it defined on the indices summed over. We also use commutativity and associativity of the addition of $V$ (Vector Space over a Field) and the order properties of Properties of the Order on the Natural Numbers.

For $n\in\mathbb{N}$ let $P(n)$ be the assertion of the lemma for that $n$: for every $b\in V^{n+1}$ and every natural number $j\le n+1$, the displayed identity holds. We prove $P(n)$ for every $n$ by induction.

Base case. Let $n=1$, let $b\in V^{2}$ and let $j\le 2$. If $j\ne2$, then $j\le1$ by claim 5 of Properties of the Order on the Natural Numbers and $1\le j$ by claim 4, so $j=1$ by claim 2; hence $j=1$ or $j=2$.

If $j=2$, then $1<j$, so $b^{(j)}_{1}=b_{1}$ and

by the recursion.

If $j=1$, then $j\le1$, so $b^{(j)}_{1}=b_{2}$ and

by commutativity of addition and the recursion. So $P(1)$ holds.

Induction step. Assume $P(n)$, let $b\in V^{n+2}$ and let $j\le n+2$, where $n+2=(n+1)+1$. By claims 4 and 5 of Properties of the Order on the Natural Numbers, either $j=n+2$ or $j\le n+1$.

Case 1: $j=n+2$. Every $k$ with $1\le k\le n+1$ satisfies $k\le n+1<n+2=j$, so $b^{(j)}_{k}=b_{k}$ for all such $k$. By the recursion,

which is the assertion for this $b$ and $j$.

Case 2: $j\le n+1$. Let $c\in V^{n+1}$ have components $c_{k}=b_{k}$ for $1\le k\le n+1$. By $P(n)$ applied to $c$ and to the index $j$,

Here $c^{(j)}_{k}=b^{(j)}_{k}$ for every $k$ with $1\le k\le n$: if $k<j$ both equal $b_{k}$, and if $j\le k$ both equal $b_{k+1}$, the index $k+1$ being at most $n+1$. Furthermore $b^{(j)}_{n+1}=b_{n+2}$, since $j\le n+1$.

Using the recursion and the restriction property,

and, using the recursion for the tuple $b^{(j)}\in V^{n+1}$,

Writing $x=\sum_{k=1}^{n}b^{(j)}_{k}$, the two right-hand sides are $(x+b_{j})+b_{n+2}$ and $(x+b_{n+2})+b_{j}$, which are equal by associativity and commutativity of the addition of $V$, both being $x+(b_{j}+b_{n+2})$. Hence the assertion holds for $b$ and $j$, and $P(n+1)$ is proved.

By induction, $P(n)$ holds for every natural number $n$.