Theorem: Vector Addition on Locally Convex Space is Continuous

Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \PP}$ be a locally convex space over $\GF$ equipped with the standard topolgoy.
Let $\struct {X \times X, \tau}$ be the Cartesian product $X \times X$ equipped with the product topology.
Let $+ : \struct {X \times X, \tau} \to \struct {X, \PP}$ be the vector addition defined on $X$.
Then $+ : \struct {X \times X, \tau} \to \struct {X, \PP}$ is a continuous mapping.

Proof:
From the definition of the standard topolgoy, the topology on $\struct {X, \PP}$ has sub-basis: 
:$\SS = \set {\map {B_p} {x, \epsilon} : p \in \PP, \, \epsilon > 0}$
where we define: 
:$\map {B_p} {x, \epsilon} = \set {y \in X : \map p {y - x} < \epsilon}$
for each $p \in \PP$ and $\epsilon > 0$.
From Continuity Test using Sub-Basis, it suffices to show that: 
:$+^{-1} \sqbrk {\map {B_p} {x_0, \epsilon} } \in \tau$ for each $p \in \PP$, $\epsilon > 0$ and $x_0 \in X$. 
That is: 
:$\set {\tuple {x, y} \in X \times X : \map p {x + y - x_0} < \epsilon} \in \tau$
For brevity, let: 
:$U_{x_0} = \set {\tuple {x, y} \in X \times X : \map p {x + y - x_0} < \epsilon}$ 
Let $\tuple {x, y} \in U_{x_0}$ and $\tuple {x', y'} \in X \times X$. 
We then have: 
{{begin-eqn}}
{{eqn	| l = \map p {x' + y' - x_0}
		| r = \map p {x' + y' - x_0 + x + y - x - y}
}}
{{eqn	| r = \map p {\paren {x + y - x_0} + \paren {x' - x} + \paren {y' - y} }
}}
{{eqn	| o = \le
		| r = \map p {x + y - x_0} + \map p {x' - x} + \map p {y' - y}
		| c = {{SeminormAxiom|3}}
}}
{{end-eqn}}
Since $\tuple {x, y} \in U_{x_0}$, we have that: 
:$\map p {x + y - x_0} < \epsilon$ 
Set: 
:$\ds \map {\epsilon'} {x, y} = \frac {\epsilon - \map p {x + y - x_0} } 2$
Then, for $\tuple {x', y'} \in \map {B_p} {x, \map {\epsilon'} {x, y} } \times \map {B_p} {y, \map {\epsilon'} {x, y} }$, we have: 
{{begin-eqn}}
{{eqn	| l = \map p {x + y - x_0} + \map p {x' - x} + \map p {y' - y}
		| o = <
		| r = \map p {x + y - x_0} + \frac {\epsilon - \map p {x + y - x_0} } 2 + \frac {\epsilon - \map p {x + y - x_0} } 2
}}
{{eqn	| r = \map p {x + y - x_0} + \epsilon - \map p {x + y - x_0}
}}
{{eqn	| r = \epsilon
}}
{{end-eqn}}
so that: 
:$\map p {x + y - x_0} < \epsilon$
and so:
:$\map {B_p} {x, \map {\epsilon'} {x, y} } \times \map {B_p} {y, \map {\epsilon'} {x, y} } \subseteq U_{x_0}$
for each $\tuple {x, y} \in U_{x_0}$.
From Union of Subsets is Subset: 
:$\ds \bigcup_{\tuple {x, y} \in U_{x_0} } \map {B_p} {x, \map {\epsilon'} {x, y} } \times \map {B_p} {y, \map {\epsilon'} {x, y} } \subseteq U_{x_0}$ 
On the other hand, for each $\tuple {x, y} \in U_{x_0}$ we clearly have: 
:$\tuple {x, y} \in \map {B_p} {x, \map {\epsilon'} {x, y} } \times \map {B_p} {y, \map {\epsilon'} {x, y} }$
So, we have: 
:$\ds U_{x_0} = \bigcup_{\tuple {x, y} \in U_{x_0} } \map {B_p} {x, \map {\epsilon'} {x, y} } \times \map {B_p} {y, \map {\epsilon'} {x, y} }$ 
From Natural Basis of Product Topology of Finite Product, we have that: 
:$\set {\map {B_p} {x, \epsilon_1} \times \map {B_q} {y, \epsilon_2} : p, q \in \PP, \, x, y \in X, \, \epsilon_1, \epsilon_2 > 0}$ 
forms a basis for $\struct {X \times X, \tau}$.
So: 
:$\map {B_p} {x, \map {\epsilon'} {x, y} } \times \map {B_p} {y, \map {\epsilon'} {x, y} } \in \tau$ for each $\tuple {x, y} \in U_{x_0}$.
Since topologies are closed under set union, we have $+^{-1} \sqbrk {\map {B_p} {x_0, \epsilon} } = U_{x_0} \in \tau$, which is what we wanted.
{{qed}}
Category:Locally Convex Spaces
Category:Vector Addition
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