Proving the triangle inequality for the discrete metric

[Terrell]

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.

 

[member 587159]

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.

Not really a lot to say. If you consider all cases, you are done. Seems like you did that, so you are fine.

 

[Terrell]

Not really a lot to say. If you consider all cases, you are done. Seems like you did that, so you are fine.

Thanks! I had it on an exam earlier, just couldn't get my head off it if I got it right or not. lol.