- #1

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## Homework Statement

Let ##x,y\in X## such that ##X## is a metric space. Let ##d(x,y)=0## if and only if ##x=y## and ##d(x,y)=1## if and only if ##x\neq y##

## Homework Equations

N/A

## The Attempt at a Solution

I have already seen various approaches in proving this. Although, I just want to know if this approach of mine is also valid.

Proof:

We want to show that ##\forall x,y,z\in X##, ##d(x,y)\leq d(x,z)+d(z,y)##.

If ##d(x,z)=1##, then ##1\leq d(x,z)+d(z,y)\leq 2##. But since ##0\leq d(x,y)\leq 1##, then ##d(x,y)\leq 1\leq d(x,z)+d(z,y)\leq 2##. However, if ##d(x,z)=0=d(z,y)##, then ##d(x,y)=0=d(x,z)+d(z,y)##. Hence, the result follows.