# Projection from Metric Space Product with Euclidean Metric is Continuous

## Theorem

Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.

Let $\AA := A_1 \times A_2$ be the cartesian product of $A_1$ and $A_2$.

Let $\MM = \struct {\AA, d}$ denote the metric space on $\AA$ where $d: \AA \to \R$ is the Euclidean metric on $\AA$:

$\map d {x, y} := \sqrt {\paren {\map {d_1} {x_1, y_1} }^2 + \paren {\map {d_2} {x_2, y_2} }^2}$

where $x = \tuple {x_1, x_2}, y = \tuple {y_1, y_2} \in \AA$.

Let $\pr_1: \MM \to M_1$ and $\pr_2: \MM \to M_2$ denote the first projection and second projection respectively on $\MM$.

Then $\pr_1$ and $\pr_2$ are both ‎continuous on $\MM$.

## Proof 1

The Euclidean metric is an instance of the $p$-product metric:

$\map {d_p} {x, y} := \paren {\paren {\map {d_1} {x_1, y_1} }^p + \paren {\map {d_2} {x_2, y_2} }^p}^{1/p}$

where $p = 2$.

The result is therefore seen to be an instance of Projection from Metric Space Product with P-Product Metric is Continuous.

$\blacksquare$

## Proof 2

We want to show that, for all $a \in \AA$:

$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall z \in \AA: \map d {z, a} < \delta \implies \map {d_1} {\map {\pr_1} z, \map {\pr_1} a} < \epsilon$

and:

$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall z \in \AA: \map d {z, a} < \delta \implies \map {d_2} {\map {\pr_2} z, \map {\pr_2} a} < \epsilon$

Let $\epsilon \in \R_{>0}$ be arbitrary.

Let $a = \tuple {x_0, y_0} \in \AA$ also be arbitrary.

Let $\delta = \epsilon$.

Let $z = \tuple {x_1, y_1} \in \AA$ such that $\map d {x, a} < \delta$.

We have:

 $\ds \map d {z, a}$ $=$ $\ds \sqrt {\paren {\map {d_1} {x_1, x_0} }^2 + \paren {\map {d_2} {y_1, y_0} }^2}$ Definition of $d$ $\ds \map {d_1} {\map {\pr_1} z, \map {\pr_1} a}$ $=$ $\ds \map {d_1} {x_1, x_0}$ Definition of First Projection $\ds \map {d_2} {\map {\pr_2} z, \map {\pr_2} a}$ $=$ $\ds \map {d_2} {y_1, y_0}$ Definition of Second Projection

Then:

 $\ds \paren {\map {d_1} {x_1, x_0} }^2 + \paren {\map {d_2} {y_1, y_0} }^2$ $\ge$ $\ds \paren {\map {d_1} {x_1, x_0} }^2$ $\text {(1)}: \quad$ $\ds \leadsto \ \$ $\ds \sqrt {\paren {\map {d_1} {x_1, x_0} }^2 + \paren {\map {d_2} {y_1, y_0} }^2}$ $\ge$ $\ds \map {d_1} {x_1, x_0}$

and similarly:

 $\ds \paren {\map {d_1} {x_1, x_0} }^2 + \paren {\map {d_2} {y_1, y_0} }^2$ $\ge$ $\ds \paren {\map {d_2} {y_1, y_0} }^2$ $\text {(2)}: \quad$ $\ds \leadsto \ \$ $\ds \sqrt {\paren {\map {d_1} {x_1, x_0} }^2 + \paren {\map {d_2} {y_1, y_0} }^2}$ $\ge$ $\ds \map {d_2} {y_1, y_0}$

Hence:

 $\ds \map d {z, a}$ $<$ $\ds \delta$ $\ds \leadsto \ \$ $\ds \sqrt {\paren {\map {d_1} {x_1, x_0} }^2 + \paren {\map {d_2} {y_1, y_0} }^2}$ $<$ $\ds \delta$ Definition of $d$ $\ds \leadsto \ \$ $\ds \map {d_1} {x_1, x_0}$ $<$ $\ds \delta$ from $(1)$ $\ds \leadsto \ \$ $\ds \map {d_1} {\map {\pr_1} z, \map {\pr_1} a}$ $<$ $\ds \epsilon$ as $\epsilon = \delta$

and:

 $\ds \map d {z, a}$ $<$ $\ds \delta$ $\ds \leadsto \ \$ $\ds \sqrt {\paren {\map {d_1} {x_1, x_0} }^2 + \paren {\map {d_2} {y_1, y_0} }^2}$ $<$ $\ds \delta$ Definition of $d$ $\ds \leadsto \ \$ $\ds \map {d_2} {y_1, y_0}$ $<$ $\ds \delta$ from $(2)$ $\ds \leadsto \ \$ $\ds \map {d_2} {\map {\pr_2} z, \map {\pr_2} a}$ $<$ $\ds \epsilon$ as $\epsilon = \delta$

We have that $a$ and $\epsilon$ are arbitrary.

Hence the result by definition of ‎continuity.

$\blacksquare$