Cartesian Space of Topological Field is Topological Vector Space

== Theorem ==
Let $K$ be a topological field.
Let $K^n$ be the cartesian $n$-th power of $K$, defining scalar multiplication and vector addition component-wise.
Equip $K^n$ with the product topology given by $K$.
Then $K^n$ is a topological vector space. 
Further, if $K$ is a Hausdorff topological field, then $K^n$ is a Hausdorff topological vector space.
== Proof ==
If $K$ is Hausdorff, then from Product of Hausdorff Factor Spaces is Hausdorff the topology on $K^n$ is Hausdorff.
We therefore just need to show that scalar multiplication and vector addition are continuous. 
=== Vector Addition is Continuous ===
Let $\pr_i^A : K^n \to K$ be the projection of $K^n$ onto its $i$th factor.
Likewise let $\pr_i^B : K^n \times K^n \to K^n$ be the projection of $K^n \times K^n$ onto its $i$th factor.
Define $f : K^n \times K^n \to K$ by: 
:$\map f {\tuple {x_1, x_2, \ldots, x_n}, \tuple {y_1, y_2, \ldots, y_n} } = \tuple {x_1 + y_1, x_2 + y_2, \ldots, x_n + y_n}$
for each $\tuple {x_1, x_2, \ldots, x_n}, \tuple {y_1, y_2, \ldots, y_n} \in K^n$.
From Continuous Mapping to Product Space, it suffices to show that $\pr_i \circ f$ is a continuous map with: 
:$\map {\paren {\pr_i^A \circ f} } {\tuple {x_1, x_2, \ldots, x_n}, \tuple {y_1, y_2, \ldots, y_n} } = x_i + y_i$ 
for each $\tuple {x_1, x_2, \ldots, x_n}, \tuple {y_1, y_2, \ldots, y_n} \in K^n$.
We can write: 
so that: 
:$\pr_i^A \circ f = \pr_i^A \circ \pr_1^B + \pr_i^A \circ \pr_2^B$
From Projection from Product Topology is Continuous, $\pr_1^B, \pr_2^B$ are continuous and: 
:$\pr_i^A$ is continuous for each $1 \le i \le n$.
So from Composite of Continuous Mappings is Continuous, we have: 
:$\pr_i^A \circ \pr_1^B$ is continuous for each $1 \le i \le n$
and:
:$\pr_i^A \circ \pr_2^B$ is continuous for each $1 \le i \le n$.
So applying Sum of Continuous Functions on Topological Ring is Continuous:
:$\pr_i^A \circ f$ is continuous for each $1 \le i \le n$.
So from Continuous Mapping to Product Space, $f$ is continuous.
=== Scalar Multiplication of Continuous ===
Again let $\pr_i^A : K^n \to K$ be the projection of $K^n$ onto its $i$th factor.
Likewise let $\pr_i^B : K \times K^n \to K^n$ be the projection of $K^n \times K^n$ onto its $i$th factor.
Define $f : K \times K^n \to K$ by: 
:$\map g {\lambda, \tuple {x_1, x_2, \ldots, x_n} } = \tuple {\lambda x_1, \lambda x_2, \ldots, \lambda x_n}$
for each $\lambda \in K, \, \tuple {x_1, x_2, \ldots, x_n} \in K^n$.
From Continuous Mapping to Product Space, it suffices to show that $\pr_i^A \circ f$ with: 
:$\map {\paren {\pr_i^A \circ g} } {\lambda, \tuple {x_1, x_2, \ldots, x_n} } = \lambda x_i$
for each $\tuple {x_1, x_2, \ldots, x_n}, \tuple {y_1, y_2, \ldots, y_n} \in K^n$, is a continuous map.
We can write:
so that: 
:$\pr_i^A \circ g = \pr_1^B \cdot \paren {\pr_i^A \circ \pr_2^B}$
From Projection from Product Topology is Continuous, $\pr_1^B, \pr_2^B$ are continuous and: 
:$\pr_i^A$ is continuous for each $1 \le i \le n$.
So from Composite of Continuous Mappings is Continuous, we have: 
:$\pr_i^A \circ \pr_2^B$ is continuous
for each $1 \le i \le n$.
Then applying Product of Continuous Functions on Topological Ring is Continuous:
:$\pr_i^A \circ g$ is continuous
for each $1 \le i \le n$.
So from Continuous Mapping to Product Space, $g$ is continuous.
Category:Topological Vector Spaces
Category:Topological Fields
Category:Hausdorff Topological Vector Spaces