Identity Mapping between Metrics separated by Scale Factor is Continuous

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be metric spaces on the same underlying set $A$.

Let $d_1$ and $d_2$ be such that:

$\forall x, y \in A: \map {d_2} {x, y} \le K \map {d_2} {x, y}$


Let $I_A: A \to A$ be the identity mapping on $A$.

Then $I_A$ is continuous from $M_1$ to $M_2$.


Proof

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

Let $a \in A$.

Set $\delta = \dfrac \epsilon K$.

Then:

\(\ds \map {d_1} {x, a}\) \(<\) \(\ds \delta\)
\(\ds \leadsto \ \ \) \(\ds \map {d_2} {\map {I_A} x, \map {I_A} a}\) \(\le\) \(\ds K \map {d_1} {x, a}\)
\(\ds \) \(<\) \(\ds K \delta\)
\(\ds \) \(=\) \(\ds \epsilon\)

Hence by definition $I_A$ is continuous at $a$.

As $a$ is arbitrary, it follows that this is true for all $a \in A$.

Thus $I_A$ is continuous on the whole of $M_1$.

$\blacksquare$


Sources