Is this a Metric Space in R x R?

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Homework Help Overview

The discussion revolves around determining whether a specific distance function defined on the Cartesian product R x R qualifies as a metric space. The distance function in question is D[(x1,y1),(x2,y2)] = min( abs(x1-x2), abs(y1-y2) ). Participants are examining the properties of this function, particularly focusing on the triangle inequality.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the triangle inequality to the distance function but expresses difficulty in manipulating the terms to fit the required form. Some participants suggest reducing the problem to known triangle inequalities in R and substituting appropriately. Others propose looking for counterexamples to aid in understanding the properties of the distance function.

Discussion Status

The discussion is active, with participants offering various suggestions and insights. Some guidance has been provided regarding the approach to take, including the use of known properties of distances in R. There are differing opinions on whether the distance function meets the criteria for being a metric, with at least one participant asserting it does not satisfy the positive definiteness property.

Contextual Notes

Participants are working within the constraints of proving the properties of a metric space, specifically focusing on the triangle inequality and positive definiteness. There is an underlying assumption that the distance function must adhere to the standard properties of metrics.

TimNguyen
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Hi.

I was trying to figure out if the following is a metric space in R x R (Cartesian product).

D[(x1,y1),(x2,y2)] = min( abs(x1-x2), abs(y1-y2) )

I know there are four properties to confirm that the following is a metric space but I'm having trouble with the "triangle inequality" for the distance function.

So, I have D[(x1,z1),(x2,z2)] should be less than or equal to D[(x1,y1),(x2,y2)] + D[(y1,z1),(y2,z2)].

D[(x1,z1),(x2,z2)] = min( abs(x1-x2), abs(z1-z2) ) and basically I'm stuck on the next step. How do I configure it such that I could put it in the form of D[(x1,y1),(x2,y2)] + D[(y1,z1),(y2,z2)]?
 
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Anybody could help...?
 
This isn't my area of expertise, but here's a suggestion: Try to reduce your problem to triangle inequalities involving distances in [itex]\mathbb{R}[/itex]. First let [itex]|x_1-x_2| \leq |y_1-y_2| \leq |z_1-z_2|[/itex]. In that case we have:

[itex]d[(x_1,y_1),(x_2,y_2)]=|x_1-x_2|[/itex]
[itex]d[(x_1,z_1),(x_2,z_2)]=|x_1-x_2|[/itex]
[itex]d[(y_1,z_1),(y_2,z_2)]=|y_1-y_2|[/itex]

At this point you can set up the triangle inequalities for the absolute values from the right hand sides, and then appropriately substitute the metric functions from the left hand sides.
 
If you're having trouble proving something, try looking for a counterexample! Even if you can't find one, the search might help you figure out the missing step of your proof.
 
One thing the triangle inequality says is that if you have a right-angled triangle, then the "length of the hypoteneuse" (by length I mean D(x,y), where x and y are the endpoints of the hypoteneuse) is less than or equal to the sum of the lengths of the other two sides of the triangle. Is this the case for your D?
 
I have an analysis book in which a distance function like the one you gave, only with max instead of min, is a metric, if that helps
.
 
It's NOT a metric, it's not positive definite. For if x1=x2 and y1 > y2, then D[(x1,y1),(x2,y2)] = 0 but (x1,y1) does not equal (x2,y2).
 
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