Finfinity is a Vector Space over F — Jemoka Knowledge Base

We define:

\begin{equation} \mathbb{F}^{\infty} = \{(x_1, x_2, \dots): x_{j} \in \mathbb{F}, \forall j=1,2,\dots\} \end{equation}

closure of addition We define addition:

\begin{equation} (x_1,x_2,\dots)+(y_1,y_2, \dots) = (x_1+y_1,x_2+y_2, \dots ) \end{equation}

Evidently, the output is also of infinite length, and as addition in \mathbb{F} is closed, then also closed. closure of scalar multiplication We define scalar multiplication:

\begin{equation} \lambda (x_1,x_2, \dots) = (\lambda x_1, \lambda x_2, \dots ) \end{equation}

ditto. as above commutativity extensible from commutativity of \mathbb{F} associativity extensible from associativity of \mathbb{F}, for both operations distribution \begin{align} \lambda ((x_1,x_2,\dots)+(y_1,y_2, \dots)) &= \lambda (x_1+y_1,x_2+y_2, \dots ) \ &= (\lambda (x_1+y_1),\lambda (x_2+y_2), \dots ) \ &= (\lambda x_1+\lambda y_1,\lambda x_2+\lambda y_2, \dots) \ &= (\lambda x_1, \lambda x_2, \dots) + (\lambda y_1, \lambda y_2, \dots) \ &= \lambda (x_1, x_2, \dots) + \lambda (y_1, y_2, \dots) \end{align} ditto. for the other direction. additive ID \begin{equation} (0,0, \dots ) \end{equation} additive inverse extensive from \mathbb{F}

\begin{equation} (-a, -b, \dots ) + (a,b, \dots ) = 0 \end{equation}

scalar multiplicative ID 1