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# Metric spaces question (4th property) watch

1. Hello there,

I am really struggling with the following question:

"Let (X,d) be a metric space. Define a function d" by

d" = min{1 , d(x,y)} for all x,y E X. Prove that d" is a metric on X" .

So far I think I have proven the first 3 properties.

(1) d"(x,y) >= 0 for all x,y E X . The proof of this is pretty straight forward.

(2)d"(x,y) = 0 iff x = y . Again the proof of this property is straight forward.

(3) d"(x,y) = d"(y,x).

For this property, I first considered the case where min{1, d(x,y)} = 1 , and said that in this case,

d"(x,y) = 1 = d"(y,x).

Then I considered the case where min{1, d(x,y)} = d(x,y) and said

d"(x,y) = d(x,y) = d(x,y) = d"(x,y).

Is this correct??

(4) the fourth property, I haven't got a clue how to verify.
(i.e. d"(x,z) <= d"(x,y) + d"(y,z) ) . I understand why this has to be true because this is just the triangle inequality. But how do I prove this property in this case???

I am really stuck. Any help would be very much appreciated.
Thank you very much for your time!
2. (Original post by beast)

(4) the fourth property, I haven't got a clue how to verify.
(i.e. d"(x,z) <= d"(x,y) + d"(y,z) ) . I understand why this has to be true because this is just the triangle inequality. But how do I prove this property in this case???

I am really stuck. Any help would be very much appreciated.
Thank you very much for your time!
[You might want to edit what you wrote for (3) - penultimate line.]

d"(x,z) <= d"(x,y) + d"(y,z)

If any the RHS are 1 then job done. If all three agree with the d-value then job done by the usual triangle inequality.

So we only need to think about when

d(x,z) > 1 and d''(x,y) = d(x,y) and d''(y,z)=d(y,z).

But then

d''(x,z) = 1 < d(x,z) <= d(x,y) + d(y,z) = d"(x,y) + d"(y,z).
3. thanks rich!!!

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