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The problem in question asks you to show that the non-negativity of a metric follows from the 2nd, 3rd, and 4th metric space axioms i.e.

$\textbf{M2}$ $d(x,y) = 0$ iff $x=y$
$\textbf{M3}$ $d(x,y) = d(y,x)$
$\textbf{M4}$ $d(x,y) \leq d(x,z) + d(z,y)$

My (kind of longwinded) proof was the following:

We prove by showing that all possible distances on the arbitrary points $x,y,z$ are non-negative.
That $d(x,x), d(y,y), d(z,z)$ are non negative is given by $\textbf{M2}$. For the remainder, let $x=y$, then using $\textbf{M2}$, $\textbf{M3}$, $\textbf{M4}$ we have

\begin{equation} \begin{split} d(x,y)=0 & \leq d(x,z) + d(z,y) \\ & \leq d(x,z) + d(z,x) \\ & \leq d(x,z) + d(x,z) \\ & \leq 2d(x,z) \end{split} \end{equation}

dividing both sides by 2 gives

\begin{equation} 0 \leq d(x,z) \end{equation}

Thus by $\textbf{M3}$ and using the fact that $y=x$ we have that $d(x,z)$, $d(z,x)$, $d(y,z)$, and $d(z,y)$ are all non-negative. QED.

I've checked some answers setting and y=x in $\textbf{M4}$ seems to be the right strategy. My question is why does setting y=x work in this case? Doesn't this only prove for the case in which y=x? What about when y doesn't equal x? I've a seen similar strategies used in proofs before and I never quite understood how prooving for a limited set of cases prooved all cases.

Many thanks.

Henry Powell
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  • It feels weird because there's no need to involve three points to begin with. Start at $x$, go to some arbitrary $y$, then go back to $x$. By the triangle inequality M4, this is a greater distance than if you just stayed at $x$, which is distance zero by M2. But now you're done, because as you correctly observed, $0 = d(x,x) \leq 2d(x,y)$. – Neal Jul 23 '20 at 17:17
  • One way to think about it is to use descriptive names for the points in the statement of M4: "start point", "intermediate point", and "end point". The use of the triangle inequality in this proof is deciding that the "start point" and "end point" are the same point, so that there are only two distinct points, the "start/end point" and the "intermediate point". (This is along the lines of what Neal said above). – Ian Jul 23 '20 at 18:38

3 Answers3

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The point is that you want to show $d(x,z) \geq 0$ for any $x,z$. This statement is only quantified over $x$ and $z$, so if you decide to introduce $y$ along the way in the proof then you can introduce it however you like. This proof picks it to be the same point as $x$, which is fine.

A different way of framing this proof: if there were some pair of points $(x,z)$ at a negative distance from one another, then the triangle inequality and symmetry would imply that $x$ is at a negative distance from itself, which contradicts M2.

Ian
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You're done, and here's why: you are trying to show that $d(x,y) \ge 0$ for all $x,y$, but you have actually shown that $d(x,z) \ge 0$ for all $x,z$. But of course, these two statements are equivalent.

You have from M2 that $d(x,x) = d(z,z) = 0$, so the $x = z$ case is covered. Now suppose $x \neq z$. You showed $0 \le d(x,z)$ by using the triangle inequality and choosing $y = x$. So $d(x,z) \ge 0$ for arbitrary $x$ and $z$. In other words, $y$ was like a dummy variable that you chose while working with arbitrary variables $x$ and $z$.

vanPelt2
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Pick some ${x}$ from your metric space. So you know that ${d(x,x)=0}$, which is an axiom. Now you also have the triangle inequality. For any three points ${x,y,z}$

$${d(x,y) \leq d(x,z) + d(z,y)}$$

The key thing is it's for any three points. Now you set ${y=x:}$

$${\Rightarrow d(x,x)\leq d(x,z) + d(z,x)}$$

Now by symmetry, because

$${d(x,z)=d(z,x)}$$

we have that

$${d(x,x)\leq 2d(x,z)\Rightarrow d(x,z)\geq d(x,x) = 0}$$

Remember, we still have not chosen what ${z}$ is. It's a free variable. It could be anything. It's some arbitrary point in the metric space. So the distance between ${x}$ and any other point $z$ has to be ${\geq 0}$. ${x}$ was also fixed at the start, but still an arbitrary choice - which proves that ${\forall\ x,z}$ in the metric space you have that

$${d(x,z)\geq 0}$$

which proves non-negativity.

Riemann'sPointyNose
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  • This is great and very clear so thanks for that. A quick follow up: is there a general way of spotting when this kind of strategy is required? I supposed it would be when you have some set of statements that involve 3 (or feasibly more?) free variables and you're only required to prove some statement that required 2? – Henry Powell Jul 23 '20 at 17:17
  • @HenryPowell No problem at all! And indeed yeah you could have noticed this way. I think another way to notice, is that we have three statements we can use - and we want to prove an inequality involving $0$. One of the statements we can use involves inequalities, and another statement involves $0$ - "so maybe we can combine these in a certain way"? :^) – Riemann'sPointyNose Jul 23 '20 at 19:10