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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.

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Welcome 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. – Kevin Carlson Oct 29 '12 at 19:16

All you need is to verify that $d(x,y)=d(y,x)$ and that $d$ is nonnegative. By the triangle inequality, $$d(x,y)+d(x,x)\geq d(y,x)$$ Subtract $d(x,y)$ from the left-hand side to see $$0=d(x,x)\geq d(y,x)-d(x,y)$$

Similarly, for non-negativity, $$d(x,y)+d(x,y)\geq d(y,y)$$ That is, $2d(x,y)\geq 0$.

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How come i do not need to show d(x,y) $\ge$ 0. – drew Oct 29 '12 at 19:16
You do. I've added that. – Kevin Carlson Oct 29 '12 at 19:20
@drew: Good question: you do. – Brian M. Scott Oct 29 '12 at 19:20
@KevinCarlson thanks for the help I get it now. – drew Oct 29 '12 at 19:25
Glad to help, @drew. As you see, you gave me an opportunity to remind myself that there are four axioms for a metric, so everybody wins! – Kevin Carlson Oct 29 '12 at 19:29

In order to show that $d$ is a metric, you must show that it is symmetric, i.e., that $d(x,y)=d(y,x)$ for all $x,y\in\bf X$, and that $d(x,y)\ge 0$ for all $x,y\in\bf X$.

For any $x,y\in\bf X$ we have $d(x,y)+d(y,x)\ge d(y,x)$, so $d(x,y)\ge 0$.

HINT: If you apply both parts of your definition to the expressions $d(x,y)+d(x,x)$ and $d(y,x)+d(y,y)$ in the right way, you can conclude that $d(x,y)=d(y,x)$; I’ll leave that for you unless you ask for more help.

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@BrianMScott thanks for the help I understand now. – drew Oct 29 '12 at 19:25
@drew: You’re welcome. – Brian M. Scott Oct 29 '12 at 19:27

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