Vector Addition on Locally Convex Space is Continuous

Theorem
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:

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:

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.