Can I use triangle inequality to divide infimum of a set?

Question

I am trying to prove $d'(x, y) = \inf\{d(x, y), 1\}$ is a metric on $X$ when $(X, d)$ is a metric space. I am struggling to prove the triangle inequality axiom.

I proved it like this:

\begin{align*} d'(x, z) + d'(z, y) &= \inf\{d(x, z), 1\} + \inf\{d(z, y), 1\} \\ &\overset{(**)}{≥} \inf\{[d(x, z) + d(z, y)], 1\} \\ &= \inf\{d(x, y), 1\} \\ &= d'(x,y) \end{align*}

However, I am not really sure if I can do the manipulation in step $(**)$. Is this wrong to assume? If not wrong, how can I prove the inequality

$$-\inf\{d(x, z), 1\} + \inf\{d(z, y), 1\} \geq \inf\{[d(x, z) + d(z, y)], 1\}$$

for a metric space?

Answer 1

You need to use the triangle inequality for the original metric $d$ also $\min\{d(x,z),1\}\leq \min\{d(x,y)+d(y,z),1\}$.

Now for (**) you just need to see that for positive $a,b,c$ $$\min\{a,c\}+\min\{b,c\} \geq \min\{a+b,c\}.$$ If you cant see this then you can write out each case, e.g.

• wlg let $a\geq c, b\leq c$, then LHS$=c+b$ and RHS$=c$.
• $a \geq c$, $b\geq c$, then LHS$=2c$ and RHS$=c$
• ...

In your case $d(x,z)=a$, $d(z,y)=b$, $c=1$.