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The following question is from Kreyszig, Introductory Functional Analysis.

$(M2) d(x,y)=0$ iff $x=y.$

$(M3) d(x,y)=d(y,x)$.

$(M4) d(x,y)\le d(x,z)+d(z,y)$.

Show that (M3) and (M4) can be obtained from (M2) and

$d(x,y)\le d(z,x)+d(z,y)$.

If I assume (M3), I can show (M4) follows from (M3) and $d(x,y)\le d(z,x)+d(z,y)$. I just can't see how to get (M3) from (M2) and $d(x,y)\le d(z,x)+d(z,y)$. Any tips?

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Try setting $z=y$ which will give one side of the inequality. Swapping variables gives the other. –  copper.hat Jan 9 '13 at 6:37

2 Answers 2

up vote 4 down vote accepted

If you choose $z$ carefully in the inequality you can get a comparison between $d(x, y)$ and $d(y, x)$. Then swapping $x$ and $y$ you can get another comparison from which you should be able to deduce $(M3)$.

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Thanks. I initially did that but it seemed to me that I had only shown it to be true for those special choices of z. Or am I missing something? –  jim Jan 9 '13 at 6:41
I'm not sure what you mean by that. There is no $z$ in $(M3)$. –  Michael Albanese Jan 9 '13 at 6:42
Also, doing it that way doesn't require using (M2). So I thought there must be way that requires (M2)? –  jim Jan 9 '13 at 6:43
I'm pretty sure you do need $(M2)$. Note that in the inequality you have two terms on the right hand side, but by making a specific choice for $z$, and using $(M2)$, you only get one. –  Michael Albanese Jan 9 '13 at 6:45
Of course, you're right (about no z term in M3)! And yes, I do remember using M2 now. Thank you very much. I understand now. –  jim Jan 9 '13 at 6:50

Set $z=y$, we get $d(x, y) \leq d(y, x) + d(y, y) = d(y, x)$. Flip $x$ and $y$ to get equality.

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Try to give a hint, not the whole idea. –  AD. Jan 9 '13 at 6:39

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