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Prove properties of distance

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,004
Hey!!

Let $v, w\in \mathbb{R} ^n$ and let $V, W\subseteq \mathbb{R} ^n$.

I want to show the following properties :
  • $d(u.,w)=0\iff u=v$
  • $d(V, W) =0\iff V\cap W\neq \emptyset$


I have done the following:

  • $d(u, w) =0\iff |u-w|=0\iff u-w=0\iff u=w$

    Or do we have to do more steps?

    $$$$
  • $d(V, W) =0\iff \min \{d(v, w) \} =0$ this means that there exists $v$ and $w$ such that $d(v, w) =0$ and from the previous one it follows that $v=w$ which means that the intersection is non empty.

    Is that correct?

(Wondering)
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,684
Hey mathmari !!

What is your function $d$ exactly? (Wondering)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,004
What is your function $d$ exactly? (Wondering)
The following definitions are given:

  • $v$ is orthogonal to $w$, if $v\cdot w=0_{\mathbb{R}}$.
  • $|v|:=\sqrt{v\cdot v}$ is the lengthh of the vector $v$.
  • $d(v,w):=|v-w|$ is the distance between the vectors $v$ and $w$.
    $d(V,W):=\min \{d(v,w)\mid v\in V, w\in W\}$ is the distance between $V$ and $W$.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,678
The following definitions are given:

  • $v$ is orthogonal to $w$, if $v\cdot w=0_{\mathbb{R}}$.
  • $|v|:=\sqrt{v\cdot v}$ is the lengthh of the vector $v$.
  • $d(v,w):=|v-w|$ is the distance between the vectors $v$ and $w$.
    $d(V,W):=\min \{d(v,w)\mid v\in V, w\in W\}$ is the distance between $V$ and $W$.
That definition of $d(V,W)$ is very suspicious because the minimum value $\min \{d(v,w)\mid v\in V, w\in W\}$ is not usually attained. For example, if you define $V, W\subset \Bbb{R}$ to be the open intervals $V=(0,1)$ and $W=(1,2)$ then $\inf \{d(v,w)\mid v\in V, w\in W\} = 0$. But $\min \{d(v,w)\mid v\in V, w\in W\}$ does not exist. You can find $v\in V$ and $w\in W$ with $d(v,w)$ arbitrarily close to $0$. But that distance can never be equal to $0$.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,684
The following definitions are given:

  • $v$ is orthogonal to $w$, if $v\cdot w=0_{\mathbb{R}}$.
  • $|v|:=\sqrt{v\cdot v}$ is the lengthh of the vector $v$.
  • $d(v,w):=|v-w|$ is the distance between the vectors $v$ and $w$.
    $d(V,W):=\min \{d(v,w)\mid v\in V, w\in W\}$ is the distance between $V$ and $W$.
With that definition I believe that your proofs are all correct. (Happy)

That definition of $d(V,W)$ is very suspicious because the minimum value $\min \{d(v,w)\mid v\in V, w\in W\}$ is not usually attained. For example, if you define $V, W\subset \Bbb{R}$ to be the open intervals $V=(0,1)$ and $W=(1,2)$ then $\inf \{d(v,w)\mid v\in V, w\in W\} = 0$. But $\min \{d(v,w)\mid v\in V, w\in W\}$ does not exist. You can find $v\in V$ and $w\in W$ with $d(v,w)$ arbitrarily close to $0$. But that distance can never be equal to $0$.
Suspicious or not, it merely means that the distance of such open intervals is not defined, isn't it?
It would probably be 'nicer' if they used $\inf$ instead of $\min$, but for the problem at hand it does not really matter does it?
Those $V$ and $W$ are probably intended to be linear or affine sub spaces or some such, in which case the distinction would not be relevant. (Thinking)
 

mathmari

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MHB Site Helper
Apr 14, 2013
4,004
With that definition I believe that your proofs are all correct. (Happy)
Can we just say that $|u-w|=0\iff u-w=0$ ? Or do we have to use some more steps in between? (Wondering)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,684
Can we just say that $|u-w|=0\iff u-w=0$ ? Or do we have to use some more steps in between?
Well, $|\cdot|$ is the Euclidean norm here.
Part of the definition of a norm, is that $|x|=0 \iff x=0$.
So as such we can simply use that. (Nod)
We might mention that it is because $|\cdot|$ is a norm.

To be fair, your $d$ is the Euclidean metric.
Part of the definition of a metric, is that $d(x,y)=0 \iff x=y$ (point separating).
So there was no actual need to prove it. We could have referred to the definition of a metric instead. (Nerd)
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,120
@mathmari

Just a note to explain why we need to be careful. The metric (distance function, norm, whatever) you are using is Eudlidean and has all the familiar properties, such as \(\displaystyle x = 0 \implies ||x|| = 0\). That's because the metric that Klaas van Aarsen mentioned (and is also the one you are using) is positive definite, which means the norm is either 0 or real.

But there are other useful metrics. For example, in SR, we have an indefinite metric, which means that \(\displaystyle x \cdot x\) can be negative, zero, or positive. (Setting c = 1 for convenience the Minkowski metric says that the vector <0, 0, 1, 0> has a norm of -1 and that the vector <1, 0, 1, 0> has a norm of 0.)

There are all sorts of fun metrics out there. (Sun)

-Dan
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
For the distance between two sets, it would be better to use "least upper bound" rather than "min". In R1 the distance between U= (0, 1) and V= (2, 3) is "lub(|x- y|, x in U, y in V)= 1" while "min(|x- y|, x in U. y in V)" does not exist.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,684
@mathmari

Just a note to explain why we need to be careful. The metric (distance function, norm, whatever) you are using is Eudlidean and has all the familiar properties, such as \(\displaystyle x = 0 \implies ||x|| = 0\). That's because the metric that Klaas van Aarsen mentioned (and is also the one you are using) is positive definite, which means the norm is either 0 or real.

But there are other useful metrics. For example, in SR, we have an indefinite metric, which means that \(\displaystyle x \cdot x\) can be negative, zero, or positive. (Setting c = 1 for convenience the Minkowski metric says that the vector <0, 0, 1, 0> has a norm of -1 and that the vector <1, 0, 1, 0> has a norm of 0.)

There are all sorts of fun metrics out there. (Sun)

-Dan
This has been a source of confusion to me. ;)

The so called Minkowski metric does not satisfy the generally used definition of a metric.
Interestingly, the wiki article of a metric does not even mention the Minkowski metric. It does mention various other flavors, such as pseudometrics, quasimetrics, and semimetrics, but the Minkowski metric is none of those!

Moreover, the wiki article for the Minkowski metric does not make any mention that it is not a metric according to the usual definition of a metric.
Confusing indeed!

Either way, as I interpret it, the Minkowski metric is formally not a metric. Instead it is a metric tensor. (Nerd)
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,120