Houjun Liu

Finfinity 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


extensible from commutativity of \(\mathbb{F}\)


extensible from associativity of \(\mathbb{F}\), for both operations


\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