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

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 $d$.

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 for $d$.

Hence is fulfilled.

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 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 is fulfilled.

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

Hence is fulfilled.

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

Then $\map d {x, y} \ne 0$ by 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 is fulfilled.

Topological Equivalence
We show that $U$ is $d$-open it is $d'$-open.

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


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




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

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 $d$.