$d_1(x,y)

Question

Can someone please give me a hint why for metric spaces we have

$d_1(x,y)<d(x,y)\Rightarrow \{x|d(x,y)<\varepsilon\}\subset \{x|d_1(x,y)<\varepsilon\}$

I have expected the opposite:

$d_1(x,y)<d(x,y)\Rightarrow \{x|d(x,y)<\varepsilon\}\supset \{x|d_1(x,y)<\varepsilon\}$

score 2 · Answer 1 · answered Sep 28 '20 at 13:13

2

Because if $d(x,y)<\varepsilon$, then $d_1(x,y)<d(x,y)<\varepsilon$.

answered Sep 28 '20 at 13:13

José Carlos Santos

427,504

score 1 · Accepted Answer · answered Sep 28 '20 at 13:17

1

The way I see it is a matter of precision, think that $x$ and $y$ are points and $d_1$ is the distance in centimeters and $d$ is in millimeters, so numerically we have $d1 <d$, and every measure using $d$ is therefore more accurate ...

answered Sep 28 '20 at 13:17

Gabriel Delgado

42

Thanks Gabriel, this goes in the right direction. Can you explain this more using $\mathbb{R}^2$. What does this mean in the space? – Averroes2 Sep 28 '20 at 13:21
Maybe this is a reason why $d$ can "see" more points? – Averroes2 Sep 28 '20 at 13:27
In the book Espaços métricos, LIMA, EL this is a matter of equivalent metrics, where one is thinner than the other, and this can occur in any set M that supports injective distance functions, that is, $(M, d) \rightarrow ( M, d_1)$ – Gabriel Delgado Sep 28 '20 at 13:51

score 0 · Answer 3 · answered Sep 28 '20 at 13:14

Yes, we claim that $\{x:d(x,y)<\varepsilon\}\subset \{x:d_1(x,y)<\varepsilon\}$. Indeed, let $x_0 \in \{x:d(x,y)<\varepsilon\}$ then, $d(x_0,y)<\varepsilon$. Since $d_1(x_0,y)<d(x_0,y)$ we obtain $d_1(x_0,y)<\varepsilon$, that is, $x_0 \in \{x: d_1(x,y)<\varepsilon\}$

$d_1(x,y)

3 Answers3