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In the Geometric Spanner Networks book by Giri Narasimhan and Michiel Smid Page 153 says

Definition 9.1.3 (Well-Separated Pair Decomposition). Let $S$ be a set of $n$ points in $\mathbb R^d$, and let $s > 0$ be a real number. A well-separated pair decomposition (WSPD) for $S$, with respect to $s$, is a sequence

$\{A_1, B_1\},\{A_2, B_2\}, \cdots , \{A_m, B_m\}$

of pairs of nonempty subsets of $S$, for some integer $m$, such that

  1. for each $i$ with $1 \leq i \leq m, A_i$ and $B_i$ are well-separated with respect to $s$, and
  2. for any two distinct points $p$ and $q$ of $S$, there is exactly one index $i$ with $1 \leq i \leq m$, such that
    • $p \in A_i$ and $q \in B_i$ or
    • $p \in B_i$ and $q \in A_i$

Look at the following image

enter image description here

The points $x_1, x_2, x_3$ are given in the plane and the split tree is drawn. Suppose for some $s$ the WSPD is built as shown. In this way the well separated pairs are going to be

$ A_1 = \{x_1, x_2\}, B_1 = \{x_3 \} $

$ A_2 = \{x_1\}, B_2 = \{x_2 \} $

The $x_1,x_2$ together are well separated to $x_3$ and $x_1$ is well separated to $x_2$. As you see $x_1 \in A_1$ and $x_1 \in A_2$, so it is in more than one pair which contradicts the definition.

I would like to know what is the problem with my example and why every point is in exactly one pair in Well Separated Pair Decomposition?

Thanks in advance.

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OK, I found it. The definition says

for any two distinct points ...

It means I must consider two points, not a one solitary. $x_1$ is well separated with $x_3$, and it is well separated with $x_2$ as well, But there is exactly one pair that separates $x_3$ from $x_1$ and also there is exactly one pair that separates $x_2$ from $x_1$

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  • $\begingroup$ Self answer is ok, if you found something that may be confusing, solved it later and moreover the question is on-topic and references are proper (everything is fine) then you should accept your answer to indicate the problem is resolved. Thank you. $\endgroup$ – Evil Nov 25 '17 at 0:31

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