Projection from Metric Space Product with Euclidean Metric is Continuous/Proof 2

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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

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.

$\blacksquare$

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