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In Munkres's topology, he proves that square metric on $\mathbb{R}^n$ is in fact a metric.

By the square metric, I mean this function:

$P:\mathbb{R}^n\times \mathbb{R}^n \rightarrow \mathbb{R}$

$P(x,y) = Max\text{{$|x_i -y_i|$}}_{1\leq i\leq n}$ where $x=(x_1,x_2,...,x_n)$ and $y=(y_1,y_2,...,y_n)$.

His proof goes as follows: The first two conditions of metric are obivious

For triangle inequality,

For any integer $i$, $|x_i-z_i|\leq|x_i-y_i|+|y_i-z_i|$ so using the definition of the function $P$, we get:

$|x_i-z_i|\leq P(x-y)+P(y-z)$

So, $P(x,z)= Max\text{{$|x_i -z_i|$}}_{1\leq i\leq n} = |x_i - z_i|\leq P(x-y)+P(y-z) $

So the triangle inequality holds.

My question is, How to move from:$|x_i-z_i|\leq |x_i-y_i|+|y_i-z_i|$ to $|x_i-z_i|\leq P(x-y)+P(y-z)$ ? How does the definition of $P$ justify that ?

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  • $\begingroup$ You switch from the notation $P(x,y)$ to $P(x-y)$. Not a big deal, but makes answering hard because we have to pick a notation. $\endgroup$ Commented Aug 1, 2014 at 16:41
  • $\begingroup$ @ThomasAndrews , I have edited it. I will follow the notation $P(x,y)$ $\endgroup$
    – FNH
    Commented Aug 1, 2014 at 17:10

1 Answer 1

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We have $|x_i-y_i|\le P(x,y)$ by definition of $P(x,y)$. Similarly, $|y_i-z_i|\leq P(y,z)$. So $$|x_i-y_i|+|y_i-z_i|\leq P(x,y)+P(y,z)$$

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