Is d(1,3) allowed to be greater than d(1,100) in metric spaces?

[kingwinner]

Let (X,d) be a metric space. d is a metric.

1) Is it possible that d(1,2)=d(1,8)?

2) Is it possible that d(1,3)>d(1,100)? If the answer is yes, wouldn't it be weird? The distance between 1 and 3 is larger than the distance between 1 and 100? This is highly counter-intuitive to me...

3) Is it possible that d(1,3)+d(3,7)≠d(1,7)?

I am very used to the usual Euclidean distance/metric, in which the above are all impossible. I'm still not entirely comfortable with the idea of a metric space. My playing around with different metrics seem to suggest that the above are possible, but it doesn't seem to make sense to me...

May someone explain this?
Thanks for any help!

 

[Redbelly98]

Moderator's note: thread moved to homework area.

The distance metric is only one example. What is the definition of a metric? Applying that definition is the way to work this problem.

 

[kingwinner]

[I'm sorry, but this is not homework. I don't think homework would ask for these kinds of problems!?]


I think the answers to all 3 questions are "yes", but it seems weird to me and I would like someone to confirm this.

Thank you!

 

[VeeEight]

Try looking at some examples of metrics and metric spaces. For example, the discrete metric can be used for one of your questions.

 

[kingwinner]

OK, I think the discrete metric gives a concrete example with d(1,2)=d(1,8), so it's possible.

How about 2) and 3)? Why are they possible? I just don't understand...It's really counter-intuitive to me...

 

[Piglet1024]

Part of the definition of a metric is the triangle inequality:

d(x,y)$$\leq$$d(x,z)+d(z,y)

Apply to number 3

 

[Dick]

OK, I think the discrete metric gives a concrete example with d(1,2)=d(1,8), so it's possible.

How about 2) and 3)? Why are they possible? I just don't understand...It's really counter-intuitive to me...

You aren't being very imaginative here. Consider a map f from your points {1,2,3,7,8} to ANY other metric space. Like R or R^2 or anything, with your choice of metric. Now define d(x,y) for x and y in R to be D(f(x),f(y)) where D is the metric in the other metric space. As long as f is injective, d is a metric on your points. Isn't it? Doesn't that make your problem easy?

 

[kingwinner]

I am still concerned with 2).

If d(1,3)>d(1,100), would this violate any of the properties of a metric?

I don't remember which metric did I get results like d(1,3)>d(1,100), I just remember that a week ago when I was playing around and computing distances between points for different metrics I got something like that which doesn't make much sense to me. I keep on telling myself that this is an abstract space, not the usual Euclidean metric, but do we actually allow things like d(1,3)>d(1,100) to happen? It's counter-intuitive and perhaps I'm lacking some imagination...

 

[Dick]

I am still concerned with 2).

If d(1,3)>d(1,100), would this violate any of the properties of a metric?

I don't remember which metric did I get results like d(1,3)>d(1,100), I just remember that a week ago when I was playing around and computing distances between points for different metrics I got something like that which doesn't make much sense to me. I keep on telling myself that this is an abstract space, not the usual Euclidean metric, but do we actually allow things like d(1,3)>d(1,100) to happen? It's counter-intuitive and perhaps I'm lacking some imagination...

It's not even all that abstract. Define f mapping your points into R with the usual metric. Define f(1)=1, f(3)=3 and f(100)=2. What's wrong with that? That's really all you need. You just are thinking of too small a class of mappings.