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This problem deals with a binary relation \( R \) defined on \( \mathbb{R}^2 \) as follows:  
\[
(x, y) R (x', y') \text{ if and only if } |x - x'| \leq y - y'.
\]

### 1. **Montrer que \( R \) est une relation d'ordre**

To show that \( R \) is an order relation, we need to check if it satisfies the three properties of a partial order:
- **Reflexivity**: For all \( (x, y) \in \mathbb{R}^2 \), we must have \( (x, y) R (x, y) \). This holds because \( |x - x| = 0 \leq y - y = 0 \), so the reflexivity property is satisfied.
  
- **Antisymmetry**: If \( (x, y) R (x', y') \) and \( (x', y') R (x, y) \), we must have \( (x, y) = (x', y') \). The condition \( (x, y) R (x', y') \) means \( |x - x'| \leq y - y' \), and \( (x', y') R (x, y) \) means \( |x' - x| \leq y' - y \). Since \( |x - x'| = |x' - x| \), the two inequalities imply \( x = x' \) and \( y = y' \), so antisymmetry holds.

- **Transitivity**: If \( (x, y) R (x', y') \) and \( (x', y') R (x'', y'') \), we must show that \( (x, y) R (x'', y'') \). From \( |x - x'| \leq y - y' \) and \( |x' - x''| \leq y' - y'' \), by the triangle inequality, \( |x - x''| \leq |x - x'| + |x' - x''| \leq (y - y') + (y' - y'') = y - y'' \), so transitivity holds.

Thus, \( R \) is a partial order.

### 2. **Dessiner les ensembles des majorants et des minorants de \( (a, b) \in \mathbb{R}^2 \)**

- **Minorants**: A point \( (x, y) \) is a minorant of \( (a, b) \) if \( |x - a| \leq y - b \). This inequality describes a "V-shaped" region below the point \( (a, b) \), where the minorants are located.
  
- **Majorants**: A point \( (x, y) \) is a majorant of \( (a, b) \) if \( |x - a| \leq b - y \). This inequality describes an inverted "V-shaped" region above the point \( (a, b) \), where the majorants are located.

### 3. **Cet ordre est-il total?**

An order is total if for every two elements \( (x, y) \) and \( (x', y') \), either \( (x, y) R (x', y') \) or \( (x', y') R (x, y) \). This is not true in this case because for certain pairs of points, neither \( |x - x'| \leq y - y' \) nor \( |x' - x| \leq y' - y \) holds. Therefore, the order is **not total**.

### 4. **Déterminer \( \sup(A) \) pour \( A = \{(x, y) \in \mathbb{R}^2 \mid x^2 + y^2 \leq 1\} \)**

The set \( A \) represents the unit disk \( x^2 + y^2 \leq 1 \), which includes all points within and on the boundary of the unit circle. To determine \( \sup(A) \) with respect to the relation \( R \), we look for the point \( (x, y) \in A \) such that \( (x, y) \) is a majorant of all other points in \( A \).

The supremum in this case would likely be the point \( (0, 1) \), as it is on the boundary of the set, and it would be able to "dominate" or be a majorant of most other points in the set \( A \) under the relation \( R \).

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Would you like a more detailed explanation on any part of this problem? Here are some further questions you might consider:

1. Can you provide an example of points for which the order is not total?
2. How does the triangle inequality help in proving transitivity in this context?
3. What would the sets of minorants and majorants look like geometrically for a specific point?
4. Can we find any other points that could serve as potential supremums of \( A \)?
5. How does the definition of the binary relation \( R \) differ from the standard Euclidean ordering?

**Tip**: Drawing a diagram of the minorant and majorant regions could help better understand the geometric aspects of the relation \( R \).

Soit R la relation binaire définie sur R^2 par: (x,y) R (x',y') si et seulement si |x - x'| ≤ y - y'. Montrez que R est une relation d'ordre, Dessiner les ensembles des majorants et des minorants de (a,b), Cet ordre est-il total?, Soit A = {(x,y) ∈ R^2 / x^2 + y^2 ≤ 1}. Déterminer sup(A).

This page explores a binary relation R on R^2 defined by |x - x'| ≤ y - y'. We prove that R is a partial order, determine its majorants and minorants, and discuss whether the order is total. Additionally, we find the supremum of a set A defined by x^2 + y^2 ≤ 1.

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