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i feel this question is over the top simple but these sort of questions screw with my thinking so i'd like clarification.

Let X be non empty and $\rho: X \times X \rightarrow \mathbb{R}$ satisfy $0 \leq \rho(x,y) < \infty~\forall x,y \in X, ~\rho(x,y)=0 \text{ iff } x=y, \text{ and } \rho(x,y) \leq \rho(x,z) + \rho(y,z) ~ \forall x,y,z \in X$ prove that $\rho$ is a metric on X

my issue comes from the fact that the question gives us that rho satsifies all the axioms of a metric on X with the exception of $d(x,y) = d(y,x)$ but that seems kind of trivial in this instance.

whats my lecturer expecting of me? is this question just an exercise in showing i know the definition of metric? or is there something im missing? if its the former then the answer would be just stating the definition of a metric, taking two elements in x and going through the axioms one at a time.

any help would be greatly appreciated. thanks.

Editted in attempt (Dead-end)

from the answer below it seems that i'm ment to show that $\rho(x,y)=\rho(y,x)$ and though i said it was trivial perhaps i jumped the gun as it seems natural but trying to prove it was a little more difficult.

heres my attempt can anyone tell me whether this is valid.

Consider the points $x,y$ then we can say that $\rho (x,y) = c_1$ from the perspective of x and $\rho (y,x)=c_2$ from the perspective of y. then the problem becomes showing that $c_1 = c_2$

Let $C$ be the set of all distances c generated from $\rho (x,y)$. C is non-empty as for $x \in X$ \rho (x,x)=0 and $0 \in C$

define $d:C\times C \rightarrow \mathbb{R}$ to be the absolute metric on C. ie $\forall c_1,c_2 \in C~, d(c_1,c_2) = |c_1-c_2|$ and so (C,d) is a metric space.

then from the definition of a metric we have $(\forall c_1,c_2 \in C \times C)[(d(c_1,c_2)=0 \Longleftrightarrow c_1 = c_2 \Rightarrow \rho(x,y)=\rho(y,x)$

does this work?

very sorry about that.

(Dead-end)

Vaas
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  • Look at your triangleinequality and set $z = x$. You obtain $\rho(x,y) \leq \rho(x,x) + \rho(y,x) = \rho(y,x)$. Interchanging the roles of $x$ and $y$ yields the result. –  Oct 01 '17 at 16:15
  • hmmm thanks for the hint i'll follow this lead next. – Vaas Oct 01 '17 at 16:26

1 Answers1

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The lecturer expects you to prove that $\rho$ is indeed a metric. So, you have to deduce from the given assumptions that $(\forall x,y\in X):\rho(x,y)=\rho(y,x)$. It should be easy for you, since you think that it seems to be “kind of trivial”.

You can prove it as follows: $\rho(x,y)\leqslant\rho(x,x)+\rho(y,x)$ (I took $z=x$), which means that $\rho(x,y)\leqslant\rho(y,x)$. For the same reason, $\rho(y,x)\leqslant\rho(x,y)$, and therefore $\rho(x,y)=\rho(y,x)$.
José Carlos Santos
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  • first. thank you for editing the original question, a bit of silliness on my part. – Vaas Oct 01 '17 at 13:28
  • @Vaas You're welcome. – José Carlos Santos Oct 01 '17 at 13:30
  • @Vaas What is $Y$? – José Carlos Santos Oct 01 '17 at 16:03
  • oh **** sorry another silly mistake on my behalf. ill edit it now. (difference between written notes and typing) they should all be C, the set of all distances from $\rho(x,y)$ – Vaas Oct 01 '17 at 16:07
  • @Vaas I don't know how is it that you go from $d(c_1,c_2)=0\iff c_1=c_2$ to $\rho(x,y)=\rho(y,x)$. Besides the fact that $c_1$ and $c_2$ have a fixed meaning here (they are $\rho(x,y)$ and $\rho(y,x)$ respectively), and therefore it makes no sense to write $(forall c_1,c_2\in C)$. – José Carlos Santos Oct 01 '17 at 16:13
  • hmmm i feel this may be my poor writing, i'll edit it to hopefully make it clearer. – Vaas Oct 01 '17 at 16:26
  • @Vaas I don't believe that any approach like that one will work. – José Carlos Santos Oct 01 '17 at 16:27
  • hmmm i think i understand why. essentially i was trying to measure the distance between distances and show that if the two distances' distance are 0 then the distances must be the same. i'll try Paul K's suggestion and use the triangle inequality.

    thanks for taking the time to look this over. i appreciate it.

     – Vaas Oct 01 '17 at 16:31
  • @Vaas I posted a solution. – José Carlos Santos Oct 01 '17 at 16:36