Pointwise Minimum of Metric and Positive Real Number is Topologically Equivalent Metric

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Theorem

Let $\struct {X, d}$ be a metric space.

Let $c > 0$ be a real number.

For each $x, y \in X$, define:

$\map {d'} {x, y} = \min \set {\map d {x, y}, c}$


Then $d'$ is a metric that is topologically equivalent to $d$.


Proof

Metric Space Axiom $(\text M 1)$

Suppose that $x, y \in X$ are such that:

$\map {d'} {x, y} = 0$

Since $c > 0$, this implies that:

$\map d {x, y} = 0$

Since $d$ is a metric, we have $x = y$ by Metric Space Axiom $(\text M 1)$ for $d$.

Hence Metric Space Axiom $(\text M 1)$ is fulfilled.

$\Box$


Metric Space Axiom $(\text M 2)$: Triangle Inequality

Let $x, y, z \in X$.

We go casewise.

Suppose that $\map d {x, y} > c$ and $\map d {y, z} > c$.

Then $\map {d'} {x, y} = c$ and $\map {d'} {y, z} = c$, so that:

$\map {d'} {x, y} + \map {d'} {y, z} = 2 c$

By definition, we also have:

$\map {d'} {x, z} \le c \le 2 c$

so that:

$\map {d'} {x, z} \le 2 c = \map {d'} {x, y} + \map {d'} {y, z}$

Now suppose that $\map d {x, y} \le c$ or $\map d {y, z} \le c$.

Then $\map {d'} {x, y} = c$ or $\map {d'} {y, z} = c$ so that:

$\map {d'} {x, y} + \map {d'} {y, z} \ge c \ge \map {d'} {x, y}$

Finally, suppose that $\map d {x, y} < c$ and $\map d {y, z} < c$.

Then, we have $\map {d'} {x, y} = \map d {x, y}$ and $\map {d'} {y, z} = \map d {y, z}$.

By Metric Space Axiom $(\text M 2)$: Triangle Inequality for $d$, we have:

$\map d {x, z} \le \map d {x, y} + \map d {y, z} = \map {d'} {x, y} + \map {d'} {y, z}$

while:

$\map {d'} {x, z} \le \map d {x, z}$

so we get:

$\map {d'} {x, z} \le \map {d'} {x, y} + \map {d'} {y, z}$

Hence Metric Space Axiom $(\text M 2)$: Triangle Inequality is fulfilled.

$\Box$


Metric Space Axiom $(\text M 3)$

For each $x, y \in X$ we have:

\(\ds \map {d'} {y, x}\) \(=\) \(\ds \min \set {\map d {y, x}, c}\)
\(\ds \) \(=\) \(\ds \min \set {\map d {x, y}, c}\) Metric Space Axiom $(\text M 3)$
\(\ds \) \(=\) \(\ds \map {d'} {x, y}\)

Hence Metric Space Axiom $(\text M 3)$ is fulfilled.

$\Box$


Metric Space Axiom $(\text M 4)$

Let $x, y \in X$ be such that $x \ne y$.

Then $\map d {x, y} \ne 0$ by Metric Space Axiom $(\text M 4)$ for $d$.

If $0 < \map d {x, y} < c$, then:

$\map {d'} {x, y} = \map d {x, y} > 0$

If $\map d {x, y} \ge c$, then:

$\map {d'} {x, y} = c > 0$

So $\map {d'} {x, y} > 0$ if $x \ne y$.

Hence Metric Space Axiom $(\text M 4)$ is fulfilled.

$\Box$


Topological Equivalence

We show that $U$ is $d$-open if and only if it is $d'$-open.

First note that if $r < c$, then for $x, y \in X$ we have:

$\map {d'} {x, y} < c$

if and only if:

$\map d {x, y} < c$

So, for each $x \in X$ and $r < c$ we have:

$\map {B_r} {x, d} = \map {B_r} {x, d'}$

Let $U$ be $d$-open.

Then, by Set is Open iff Union of Open Balls, for each $x \in U$, there exists $r_x > 0$ such that:

$\ds U = \bigcup_{x \mathop \in U} \map {B_{r_x} } {x, d}$

Setting:

$\delta_x = \min \set {r_x, \dfrac c 2}$

we have:

$\map {B_{\delta_x} } {x, d} \subseteq \map {B_{r_x} } {x, d}$

so that:

$\ds \bigcup_{x \mathop \in U} \map {B_{\delta_x} } {x, d} \subseteq \bigcup_{x \mathop \in U} \map {B_{r_x} } {x, d} \subseteq U$

Conversely, $x \in \map {B_{\delta_x} } {x, d}$ for each $x \in U$, so that:

$\ds U \subseteq \bigcup_{x \mathop \in U} \map {B_{\delta_x} } {x, d}$

giving:

$\ds U = \bigcup_{x \mathop \in U} \map {B_{\delta_x} } {x, d}$

Since $\delta_x < c$, we have:

$\ds U = \bigcup_{x \mathop \in U} \map {B_{\delta_x} } {x, d'}$

So $U$ is $d'$-open.

Swapping $d$ and $d'$ in the above proof, we get that if $U$ is $d'$-open, it is $d$-open.

So $d'$ is a metric that is topologically equivalent to $d$.

$\blacksquare$

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