For $x,y,z \in \mathbb{R}^n$ is it true that $\max\{|x_i-z_i|\} \leq \max\{|x_i-y_i|\} + \max\{|y_i-z_i|\}$ for $1 \leq i \leq n$

general-topologymetric-spacessolution-verification

For $x,y,z \in \mathbb{R}^n$ is it true that $\max\{|x_i-z_i|\} \leq \max\{|x_i-y_i|\} + \max\{|y_i-z_i|\}$ for $1 \leq i \leq n$?

Here I want to use the metric $d(x,y)=|x_1-y_1|+…+|x_n-y_n|$ to prove that the Triangle inequality holds for $d_{max}=\max\{|x_i-y_i|: 1 \leq i \leq n\}$.

Since $d$ is a metric on $\mathbb{R}^n$ we have that for any $x,y,z \in \mathbb{R}^n$, $d(x,z)\leq d(x,y)+d(y,z)$. Then for $1 \leq i \leq n$ we have,
$$|x_i-z_i| \leq |x_i-y_i|+|y_i-z_i|$$

I am having trouble explicitly linking this statement to the desired result. In other words, precisely why does this inequality still hold when we take the maximum of the sets over $i$?

Best Answer

For each $i$ thanks to the triangular inequality you have $$|x_{i}-z_{i}| \leq |x_{i}-y_{i}|+|y_{i}-z_{i}|.$$ Now, note that $|x_{i}-y_{i}| \leq \max \{ |x_{i}-y_{i}| \}$, and $|y_{i}-z_{i}| \leq \max \{ |y_{i}-z_{i}| \}$. Hence, $$ |x_{i}-z_{i}| \leq \max \{ |x_{i}-y_{i}| \} +\max \{ |y_{i}-z_{i}| \}.$$ Now take the maximum in this expression to obtain the desire inequality.

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