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

Let us analyze the properties of the relation $\preccurlyeq$ on $\mathcal{F}$:

1. Reflexivity: For any $f \in \mathcal{F}$, since $f(x) \le f(x)$ for all $x \in \{1, \dots, n\}$, we have $f \preccurlyeq f$.

2. Antisymmetry: If $f \preccurlyeq g$ and $g \preccurlyeq f$, then for every $x$, $f(x) \le g(x)$ and $g(x) \le f(x)$, which implies $f(x) = g(x)$ for all $x$, so $f = g$.

3. Transitivity: If $f \preccurlyeq g$ and $g \preccurlyeq h$, then for all $x$, $f(x) \le g(x) \le h(x) \implies f(x) \le h(x)$, so $f \preccurlyeq h$.

Since $\preccurlyeq$ is reflexive, antisymmetric, and transitive, $(\mathcal{F}, \preccurlyeq)$ is a partial order (poset). Hence, (B) is TRUE, and (A) and (D) are FALSE (because a non-trivial partial order is antisymmetric, not symmetric or an equivalence relation).

4. Lattice property: For any two functions $f, g \in \mathcal{F}$, we can define:

   • Join (least upper bound):$(f \lor g)(x) = \max(f(x), g(x))$

   • Meet (greatest lower bound):$(f \land g)(x) = \min(f(x), g(x))$

   Both $f \lor g$ and $f \land g$ are valid functions in $\mathcal{F}$. Since every pair of elements has a unique least upper bound and greatest lower bound, $(\mathcal{F}, \preccurlyeq)$ is a lattice (in fact, isomorphic to the Boolean algebra $\{0,1\}^n$ / power set lattice $(\mathcal{P}(\{1, \dots, n\}), \subseteq)$).

Therefore, statements (B) and (C) are TRUE.