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Can anyone explain the second bullet point on the following answer?

I am trying to understand how this inequality is proven:

$$\delta(A,B)=|A|+|B|=|A|+|B-C+C|\leq |A|+|C|+|B-C|=\delta(A,C)+\delta(C,B)$$

How does one conclude that the inequality is correct and also how are these absolute values manipulated? If anyone can explain this step by step please.

Michael Munta
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1 Answers1

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First of all a useful zero is filled in and then a simple application of the triangle inequality follows which is $$|x + y| \le |x| +|y|$$

In more simple steps: $$\begin{align} |B| &= |B + 0| \\ &= |B + (C-C)| \\ &= |(B-C) + C| \\ &\le |B-C| + |C|\end{align}$$

So it follows: $$|A| + |B| \le |A| + |B-C| + |C|$$

Gono
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  • What bothers me is the fact that he defined the metric space as $\mathbb{R}^2$ and then defined the metric as if it were $\mathbb{R}^1$. Is that a mistake? – Michael Munta Mar 06 '20 at 20:22
  • No, the definition $$\delta(A,B)=\begin{cases} |A|+|B|, & \text{if $A\neq \lambda B$} \ |A-B|, & \text{if $A= \lambda B$ for some $\lambda$} \end{cases}$$ is well defined for $A,B \in \Bbb R^n$. If you are confused about the absolute value: It's a short notation for the euclidean norm, so $$|\cdot| = ||\cdot||_2$$ Maybe if we write the metric as $$\delta(A,B)=\begin{cases} ||A||_2+||B||_2, & \text{if $A\neq \lambda B$} \ ||A-B||_2, & \text{if $A= \lambda B$ for some $\lambda$} \end{cases}$$ is more familiar to you? – Gono Mar 07 '20 at 07:06