Let F be the set of all functions from {1,..., n} to {0,1}. Define the binary relation ≤ on F as follows:
∀f,g ∈ F, f ≤ g if and only if ∀ x ∈ {1, ..., n}, f(x) ≤ g(x), where 0 ≤ 1.
Which of the following statement(s) is/are TRUE?

The correct options are:
1. (F, ≤) is a partial order
2. (F, ≤) is a lattice

1. Proof that (F, ≤) is a partial order:
A binary relation ≤ on a set F is a partial order if it is reflexive, antisymmetric, and transitive.

Reflexivity:
Let $f\in F$. For every $x\in \{1,...,n\}$, $f\left(x\right)=f\left(x\right)$. Since the standard ordering on $\{0,1\}$ is reflexive, we have:
$f\left(x\right)\le f\left(x\right)$ for all $x\in \{1,...,n\}$.
By definition, this implies $f\le f$. Thus, the relation is reflexive.

Antisymmetry:
Let $f,g\in F$ such that $f\le g$ and $g\le f$.
By definition:
For all $x\in \{1,...,n\}$, $f\left(x\right)\le g\left(x\right)$ and $g\left(x\right)\le f\left(x\right)$.
Since the relation ≤ on the set of codomain values $\{0,1\}$ is antisymmetric, this implies:
$f\left(x\right)=g\left(x\right)$ for all $x\in \{1,...,n\}$.
Therefore, $f=g$. Thus, the relation is antisymmetric.

Transitivity:
Let $f,g,h\in F$ such that $f\le g$ and $g\le h$.
By definition:
For all $x\in \{1,...,n\}$, $f\left(x\right)\le g\left(x\right)$ and $g\left(x\right)\le h\left(x\right)$.
Since the standard relation ≤ on $\{0,1\}$ is transitive, it follows that:
$f\left(x\right)\le h\left(x\right)$ for all $x\in \{1,...,n\}$.
By definition, this implies $f\le h$. Thus, the relation is transitive.

Since ≤ is reflexive, antisymmetric, and transitive, $(F,\le )$ is a partially ordered set (poset).

2. Proof that (F, ≤) is a lattice:
A partially ordered set is a lattice if every pair of elements has a unique supremum (least upper bound, LUB) and a unique infimum (greatest lower bound, GLB).
Let $f,g\in F$.

Define a function $h\in F$ pointwise as:
$h\left(x\right)=max\left(f\right(x),g(x\left)\right)$ for all $x\in \{1,...,n\}$.
Since the codomain of both $f$ and $g$ is $\{0,1\}$, the maximum at each point is also in $\{0,1\}$, so $h\in F$.
This function $h$ acts as the least upper bound (join) of $f$ and $g$ because:
- $f\left(x\right)\le h\left(x\right)$ and $g\left(x\right)\le h\left(x\right)$ for all $x$, which means $f\le h$ and $g\le h$ (so $h$ is an upper bound).
- If there is any other upper bound $k\in F$ such that $f\le k$ and $g\le k$, then for all $x$, $f\left(x\right)\le k\left(x\right)$ and $g\left(x\right)\le k\left(x\right)$, which implies $max\left(f\right(x),g(x\left)\right)\le k\left(x\right)$, meaning $h\le k$. Thus, $h$ is the least upper bound, written as $f\vee g$.

Similarly, define a function $u\in F$ pointwise as:
$u\left(x\right)=min\left(f\right(x),g(x\left)\right)$ for all $x\in \{1,...,n\}$.
By a symmetric argument, $u$ is the greatest lower bound (meet) of $f$ and $g$, written as $f\wedge g$.
Since every pair of functions has a unique join and meet, $(F,\le )$ is a lattice.

3. Why the other options are incorrect:
- ≤ is a symmetric relation: This is false. Consider the constant functions $f\left(x\right)=0$ and $g\left(x\right)=1$ for all $x$. Since $0\le 1$ at all points, we have $f\le g$. However, $g\nleqq f$ because $1\nleqq 0$.
- ≤ is an equivalence relation: Since an equivalence relation must be symmetric and this relation is not, ≤ is not an equivalence relation.