How to Understand the Standard Bounded Metric

Question

I would like to check if my following understanding of the standard bounded metric from Topology written by James R. Munkres is correct or not:

The standard bounded metric is defined by Theorem 20.1 on the page 121 as follows:

Let $\mathbf X$ be a metric space with metric $\mathtt d$. Define $$\bar{d}(x,y) = \min\{d(x,y),1\}$$.

Then $\bar{d}$ is a metric called the standard bounded metric.

Based on the above conception, can we interpret $\bar{d}$as follows:

1. If $d(x,y)<1$, then $\bar{d}=d(x,y)$, meaning we choose $d(x,y)$ not "1" in $\min\{d(x,y),1\}$ ?
2. If $d(x,y)\ge1$, then $\bar{d}=1$, meaning we choose "1" not $d(x,y)$ in $\min\{d(x,y),1\}$ ?
3. In any situation, $\bar{d}$ does not exceed the value one, meaning $\bar{d}\le1$ holds forever?

Answer 1

The idea is that metric is concerned with things "very close" rather than "very far". So if you define $\bar d$ as you do, you don't lose a whole lot of information.

This is like looking at the horizon. You will have the same information about what you can see even if you treat everything past the horizon as the same distance from you.

So if I look outside, from my point of view England and Brazil are both the same distance from me, they are beyond the horizon. On the other hand, the next block is further away from the tree outside my window.

The definition of $\bar d$ is the same. It takes $d$ and sets the horizon to $1$. So everything closer than $1$ unit of distance is measured as it were before, and everything further than $1$ is just cut off there. So indeed the metric is bounded by $1$.

Answer 2

Everything you said is correct. The first two follow from the definition of minimum and the last one is a consequence of the definition of minimum.

Answer 3

1) If $d(x,y)=a<1$ then $\overline{d}=\min\{a,1\}=a$.

2) If $d(x,y)=b\geq 1$ then $\overline{d}=\min\{b,1\}=1$.

3) For any value of $d(x,y)$, we always have $\overline{d}\leq 1$.

Answer 4

@AsafKaragila Impressive. I would like to extend my understanding from two sides:

1. If we extend $\bar{d}(x,y) = \min\{d(x,y),1\}$ from $\mathbb R^1$ to $\mathbb R^\omega$, then $\bar{d}(\mathbf x,\mathbf y) = \min\{d(x_\alpha,y_\alpha),1\}$, $\alpha\in\omega$. Can we interpret it as this example:

   $\bar{d}(\mathbf x,\mathbf y) = \min\{d(x_1,y_1),1\}$, when $\alpha=1$;

   $\bar{d}(\mathbf x,\mathbf y) = \min\{d(x_2,y_2),1\}$, when $\alpha=2$;

   $\bar{d}(\mathbf x,\mathbf y) = \min\{d(x_3,y_3),1\}$, when $\alpha=3$;

   ......

   $\bar{d}(\mathbf x,\mathbf y) = \min\{d(x_\alpha,y_\alpha),1\}$

   ......

   And if the above example is correct, $\bar{d}(\mathbf x,\mathbf y) ={d(x_\alpha,y_\alpha)}$, when ${d(x_\alpha,y_\alpha)}<1$. I think the above example is correct.

2. Considering the uniform metric $\bar{\mathbf \rho}$ on $\mathbb R^\omega$, Can we interpret the metric as the following example:

   $\bar{\mathbf \rho} = \sup\{\bar{d}(\mathbf x_1,\mathbf y_1)\}$, if $\omega=1$;

   $\bar{\mathbf \rho} = \sup\{\bar{d}(\mathbf x_1,\mathbf y_1)\ , \bar{d}(\mathbf x_2,\mathbf y_2)\}$, if $\omega=2$;

   $\bar{\mathbf \rho} = \sup\{\bar{d}(\mathbf x_1,\mathbf y_1)\ , \bar{d}(\mathbf x_2,\mathbf y_2)\ , \bar{d}(\mathbf x_3,\mathbf y_3)\}$, if $\omega=3$;

   ......

   $\bar{\mathbf \rho} = \sup\{\bar{d}(\mathbf x_\alpha,\mathbf y_\alpha)\}$, if $\alpha\in\omega$;

   Meaning $\bar{\mathbf \rho}$ is not less than the maximum of the elements in the set $\{\bar{d}(\mathbf x_1,\mathbf y_1)\ , \bar{d}(\mathbf x_2,\mathbf y_2)\ , \bar{d}(\mathbf x_3,\mathbf y_3)\ , ......, \bar{d}(\mathbf x_\alpha,\mathbf y_\alpha)\ , ......\}$, when we pick up the largest element from the set to compare with $\bar{\mathbf \rho}$ ?

I apologize for asking you by the way of answering my own question, 'cause I found that it was too long by 1255 characters when I wanted to add my comment on your answer......