Question

Let \( \mathcal{F} \) be the set of all functions from \(\{1, \dots, n\}\) to \(\{0,1\}\). Define the binary relation \(\preceq\) on \( \mathcal{F} \) as follows: \[ \forall f, g \in \mathcal{F}, \quad f \preceq g { if and only if } \forall x \in \{1, \dots, n\}, \quad f(x) \leq g(x), \quad {where } 0 \leq 1. \] Which of the following statement(s) is/are TRUE?

• A. \(\preceq\) is a symmetric relation
• B. \( (\mathcal{F}, \preceq) \) is a partial order
• C. \( (\mathcal{F}, \preceq) \) is a lattice
• D. \(\preceq\) is an equivalence relation

Hint:

{A partial order defines a hierarchical structure, whereas a lattice ensures the existence of least upper bounds and greatest lower bounds.}

Answer 1

The Correct Option is B, C

- The relation \( \preceq \) is not symmetric, since \( f \preceq g \) does not imply \( g \preceq f \) unless \( f = g \). Symmetry would require that if \( f \) is related to \( g \), then \( g \) must also be related to \( f \). However, in this case, \( f \preceq g \) simply means that \( f \) is less than or equal to \( g \) according to the ordering, and this is not necessarily true in reverse unless \( f = g \).
- The relation is a partial order because it satisfies:
• Reflexivity: \( f \preceq f \) for all \( f \in \mathcal{F} \). Reflexivity means that any element is related to itself.
• Antisymmetry: If \( f \preceq g \) and \( g \preceq f \), then \( f = g \). Antisymmetry ensures that if two elements are related in both directions, they must be equal.
• Transitivity: If \( f \preceq g \) and \( g \preceq h \), then \( f \preceq h \). Transitivity means that if one element is related to another, and that second element is related to a third, the first element is related to the third.
- The structure \( (\mathcal{F}, \preceq) \) forms a lattice since every pair of functions has a unique least upper bound (pointwise maximum) and greatest lower bound (pointwise minimum). In a lattice, the least upper bound and greatest lower bound can always be found for any two elements in the set.
- However, \( \preceq \) is not an equivalence relation, since it lacks symmetry. An equivalence relation requires symmetry, which is not satisfied in this case because \( f \preceq g \) does not imply \( g \preceq f \), unless \( f = g \).
Thus, the correct options are (B) and (C).