If d is a real-valued function on $\mathbf X \times \mathbf X$ which for all x,y and z in $\mathbf X$ satisfies $$ d(x,y)=0 \iff x=y $$ $$ d(x,y)+d(x,z) \ge d(y,z) $$ show that d is a metric. Problem comes from Real analysis Haaser and Sullivan.
Weakened criteria for a metric proof.
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real-analysis
general-topology
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2Welcome to Math StackExchange, drew. For future questions, especially ones which might require more work to solve, it's very much appreciated if you'll include some comments about what you've tried on your own before coming here here to ask for help. – 2012-10-29