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Real analysis metric spaces

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7. If d is a real-valued function on which for all x, y, and z in X satistifes
d(x,y) = 0 if and only if x=y
d(x,y)+d(x,z)≥d(y,z)
show that d is a metric on X.

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This solution is comprised of a detailed explanation to show that d is a metric on X.

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Proof:
(1) d(x,y)>=0
Since d(x,y)+d(x,z)>=d(y,z), especially, we select z=y and from the condition, we know d(z,z)=0, then we have d(x,y)+d(x,y)>=d(z,z)=0, ...

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