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I am reading about Monotone Comparative Statics in Levin's lecture notes. At page 35 there is this picture, which provides an example of two subsets of $\mathbb{R}$. By the usual definition (Tokpins, 1998):

For two sets of real numbers A and B, define the binary relation $\leq_{s}$ as follows: $A \leq_{s} B \quad $ if for any $a \in A$ and $ b \in B$, $min\{a,b\} \in A$ and $max\{a,b\} \in B$, so that A is greater than or equal to B in the strong set order.

In the notes the first example $A\leq_s B$ while in the second case this is not true. However, I cannot understand the concept. Any idea?

enter image description here

Mino
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  • These appear to be the same example - how do the sets $A$ and $B$ differ from one figure to the next, otherwise? – Chris Jun 22 '17 at 17:28
  • In the second one only the singletons are part of the set I believe – Mino Jun 22 '17 at 17:30
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    Note ${1,2,3,4}$ the points in the picture. We then have $A={1,2,3}$ and $B={2,4}$. We have $3\in{A}$ and $2\in{B}$, but $\max{{2,3}}=3\notin{B}$. – matboy Jun 22 '17 at 17:31
  • Exactly, but then why this reasoning does not apply to the first picture? – Mino Jun 22 '17 at 17:39

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The relation $ A \leq_s B$ states that $ A $ is a beginning piece of $ A\cup B $ and that $ B$ is an end piece of $ A\cup B$. You see that in t he first example this is the case. However in the second example, $ A$ is a beginning piece of $ A \cup B $, but $ B $ is not an end piece of $ A\cup B$, since the third point belongs to $ A$ but not to $ B$, anyhow the second point is in $ B $

Andreas Lietz
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  • exactly, I realised it today: I don't know why I thought it should be mutually exclusive, why is clearly not. Although it is trivial, I will not delete the topic, since I haven't found another talking about the issue – Mino Jun 23 '17 at 07:30