Question

Find all the metric on set M consisting of two points.

Answer 1

You mean all of the metrics?

Answer 2

yes

Answer 3

Well, let the set by S = { a, b}
Whatever the metric is, we must have d(a,b) > 0 and d(a,a) = d(b,b) = 0

Now, how can you define such a metric?

Answer 4

...and d(a,b) = d(b,a). All of these constraints on the metric d are consequences of the axioms of a metric space.

Answer 5

wat about triangle inequality

Answer 6

There are not three elements, so the triangle inequality axiom is moot

Answer 7

...and it will trivially be the case if you have defined d correctly that
\[ d(a,b) \leq d(a,a) + d(a,b), \ \ etc. \]

Answer 8

@JamesJ nice to see you again
i have a question
is it \(d(a,b)>0\) if \(a\neq b\) or is it \(d(a,b)\geq 0\) so you can have the zero metric?

Answer 9

Hi Sat, you too.

I'll ask walters what he is using, but the standard axiom is d(x,y) = 0 if and only if x = y.

Answer 10

k thanks!

Answer 11

i am using d(a,b)>=0

Answer 12

it means by symmetry i will have

d(a,b)=d(b,a)

Answer 13

Yes.

So in summary, what is a definition of d that works here?

Answer 14

d is a metric

Answer 15

You need to give an explicit definition of d, the function
\[ d : S \times S \rightarrow \mathbb{R}_{\geq 0} \]

Answer 16

d(a,a) = ...
d(a,b) = ...
d(b,a) = ...
d(b,b) = ....

Answer 17

so in this case is that all

Answer 18

You need to give the EXPLICIT definition of what d is. What explicitly are each of these terms equal to, what real numbers?
d(a,a) = ...
d(a,b) = ...
d(b,a) = ...
d(b,b) = ....

Answer 19

i don't get u the statement u wrote

Answer 20

What is a metric space? A metric space is a set S together with a function
\[ d : S \times S \rightarrow \mathbb{R}_{\geq 0} \]
where d follows the three metric set axioms
1) d(x,y) = 0 if and only if x = y
2) for all x, y in S, d(x,y) = d(y,x)
3) for all x, y, z in S, \( d(x,y) \leq d(x,z) + d(z,y) \)

Your set contains two elements, S = { a, b }

To give a metric on this set, you need to the define the function d.

That is, you need to write down explicitly the rule that associates every member of the domain S x S with a value in the co-domain, the non-negative real numbers.

S x S = { (a,a), (a,b), (b,a), (b,b) }

Hence to specify a metric d, you need to say explicitly what unique real number is assigned to each of
d(a,a)
d(a,b)
d(b,a)
and d(b,b)

Answer 21

*correction: where d follows the three metric SPACE axioms

Answer 22

Note the the question implies there is more than one possible solution for d here, so don't necessarily expect to find a unique solution for d.

Answer 23

Can you think of at least one, or just one, definition of d?

Answer 24

jacobian, please don't give the answer

Answer 25

walters, talk to me. can't read your mind over these damn inter-web thingies. ;-)

Answer 26

d:R x R->R

Answer 27

? What? That's not a definition of d in this case at all.

Answer 28

Let me ask you this. Would this be a valid metric on S?

d(a,a) = 1
d(a,b) = 2
d(b,a) = 3
d(b,b) = 4

Why or why not?

Answer 29

no because it does not satisfy the all the axioms

Answer 30

Right.

So write down a definition of d that does meet the axioms.

Answer 31

(a,a) = 0
d(a,b) = 0
d(b,a) = 0
d(b,b) = 0

Answer 32

No, because d(a,b) and d(b,a) can't be zero.

Answer 33

d(a,a) = 0
d(a,b) = x (where x >0
d(b,a) = x (where x>0)
d(b,b) = 0

Answer 34

Yes, for any x > 0, what you have written down is a metric.

Answer 35

ok now i understand , it means it has answered the question

Answer 36

yes

Answer 37

wow thnx a bunch