Math Problem Statement

We define the relation ⪯ by the set R^2 by setting (x, y) ⪯(a, b) if x ≤ a and y ≤ b. Is ⪯ a partial order on the set R^2? What if it is arrangement?

Solution

To analyze whether the relation \preceq defined on the set R2\mathbb{R}^2 is a partial order or an arrangement, let us examine the definition and its properties step by step:

The relation is defined as: (x,y)(a,b)    xa and yb.(x, y) \preceq (a, b) \iff x \leq a \text{ and } y \leq b.

1. Partial Order

A partial order must satisfy the following three properties:

  1. Reflexivity: (x,y)(x,y)(x, y) \preceq (x, y) for all (x,y)R2(x, y) \in \mathbb{R}^2.
  2. Antisymmetry: If (x,y)(a,b)(x, y) \preceq (a, b) and (a,b)(x,y)(a, b) \preceq (x, y), then (x,y)=(a,b)(x, y) = (a, b).
  3. Transitivity: If (x,y)(a,b)(x, y) \preceq (a, b) and (a,b)(c,d)(a, b) \preceq (c, d), then (x,y)(c,d)(x, y) \preceq (c, d).

Reflexivity:

For any (x,y)R2(x, y) \in \mathbb{R}^2, xxx \leq x and yyy \leq y are true. Hence, (x,y)(x,y)(x, y) \preceq (x, y). Reflexivity holds.

Antisymmetry:

If (x,y)(a,b)(x, y) \preceq (a, b), then xax \leq a and yby \leq b. If (a,b)(x,y)(a, b) \preceq (x, y), then axa \leq x and byb \leq y. Combining these, x=ax = a and y=by = b. Thus, (x,y)=(a,b)(x, y) = (a, b). Antisymmetry holds.

Transitivity:

If (x,y)(a,b)(x, y) \preceq (a, b), then xax \leq a and yby \leq b. If (a,b)(c,d)(a, b) \preceq (c, d), then aca \leq c and bdb \leq d. Combining these, xcx \leq c and ydy \leq d, so (x,y)(c,d)(x, y) \preceq (c, d). Transitivity holds.

Thus, the relation \preceq is a partial order on R2\mathbb{R}^2.


2. Arrangement

An arrangement (or total order) requires the relation to be a partial order and satisfy comparability, meaning for any two elements (x,y),(a,b)R2(x, y), (a, b) \in \mathbb{R}^2, either: (x,y)(a,b)or(a,b)(x,y).(x, y) \preceq (a, b) \quad \text{or} \quad (a, b) \preceq (x, y).

In this case, comparability does not hold because it is possible to have two pairs (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) such that neither (x1,y1)(x2,y2)(x_1, y_1) \preceq (x_2, y_2) nor (x2,y2)(x1,y1)(x_2, y_2) \preceq (x_1, y_1). For example: (1,2)(2,1)and(2,1)(1,2),(1, 2) \npreceq (2, 1) \quad \text{and} \quad (2, 1) \npreceq (1, 2), because 1≰21 \not\leq 2 and 2≰12 \not\leq 1.

Hence, \preceq is not an arrangement.


Conclusion:

The relation \preceq is a partial order on R2\mathbb{R}^2, but it is not an arrangement.


Follow-Up Questions:

  1. Can you find an example of a different partial order on R2\mathbb{R}^2?
  2. How does the concept of a lattice relate to this partial order?
  3. What modifications to \preceq would make it a total order?
  4. Can we extend this relation to higher dimensions, e.g., Rn\mathbb{R}^n?
  5. How does this partial order compare to lexicographical ordering on R2\mathbb{R}^2?

Tip: Always verify all three properties (reflexivity, antisymmetry, and transitivity) when evaluating partial orders.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Partial Order
Relations on Sets
Order Theory

Formulas

(x, y) ⪯ (a, b) if and only if x ≤ a and y ≤ b

Theorems

Reflexivity, Antisymmetry, Transitivity for Partial Orders

Suitable Grade Level

Undergraduate - Discrete Mathematics