---
title: "Finfinity is a Vector Space over F"
source: https://www.jemoka.com/posts/kbhfinfty_is_a_vector_space_over_f/
---

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)) &amp;= \lambda (x_1+y_1,x_2+y_2, \dots ) \\ &amp;= (\lambda (x_1+y_1),\lambda (x_2+y_2), \dots ) \\ &amp;= (\lambda x_1+\lambda y_1,\lambda x_2+\lambda y_2, \dots) \\ &amp;= (\lambda x_1, \lambda x_2, \dots) + (\lambda y_1, \lambda y_2, \dots) \\ &amp;= \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 \(1\)
