- #1

- 5

- 0

You should upgrade or use an alternative browser.

- Thread starter ag2ie
- Start date

- #1

- 5

- 0

- #2

- 22,089

- 3,296

Well, what do you think?? What have you tried?

- #3

- 5

- 0

It's easy to verify d(x,y)=0 iff x=y and d(x, y)=d(y,x),

but I don't know how to prove triangle Inequality...

- #4

- 22,089

- 3,296

Well, is the following trye

[tex]d(x,z)\leq \max\{d(x,y),r(x,y)\}+\max\{d(y,z),r(y,z)\}[/tex]

???

Last edited:

- #5

- 5

- 0

yes..but if r(x,z) is greater than d(x,z), and r(y,z) is smaller than d(y,z), then this inequality is not necessary true...right?

- #6

- 22,089

- 3,296

Sorry, I made a typo, check the post again.

- #7

- 5

- 0

I see...if r(x,y) is greater than d( x,y), then d(x,z)≤max{d(x,y),r(x,y)}+max{d(y,z),r(y,z)} is also true...

Thanks ...and I think (X, min(d, r)) is not a metric space..right?

- #8

- 22,089

- 3,296

- #9

- 5

- 0

Thanks..you are really helpful

Share: