Prove anti-symmetric-ness of partial ordered set in lattice.

Question

Prove anti-symmetric-ness of partial ordered set in lattice.

Definition:

If $(A, \le_{A})$ is a lattice and $C$ is a set, $([C \rightarrow A], \le)$ is also a lattice.

And $\rightarrow$ is defined as follows: $f \le g$ if and only if for any $c \in C$, $f(c) \le_{A} g(c)$

Such lattices are defined as point wise.

And for $([C \rightarrow A], \le)$ as point wise lattice, we can drop $_{A}$.

Then I need to prove that $\le$ of $([C \rightarrow A], \le)$ is anti-symmetric.

And below is my proof based on this posting.

1. Let $f, g \in [C \to A]$ be arbitrary
2. such that $f \leq g$ and $g \leq f$.
3. Then by definition $f(c) \leq_A g(c)$ 4, and $g(c) \leq_A f(c)$ for any $c \in C$.
4. Since $A$ is a lattice, it is a partially ordered set
5. , therefore anti-symmetric.
6. We have $f(c), g(c) \in A$
7. We have $f(c) \leq_A g(c)$ and $g(c) \leq_A f(c)$
8. , thus $f(c) = g(c)$ for any $c \in C$ by the anti-symmetric-ness of $A$
9. This means $f = g$
10. Therefore, $\le$ of $([C \rightarrow A], \le)$ is anti-symmetric.

Is it correct?

Thanks in advance.

Answer 1

In this answer I assume $[C\to A]$ denotes the set of functions from $C$ to $A$.

Note that if $C$ is the empty set, then $[C\to A]=\varnothing$ and thus $\left([C\to A],\leq \right)$ can't be a lattice. Despite this, $\leq$ comes out trivially anti-symmetric.

The rest is essentially correct. On step 8. the justification should be anti-simmetry instead of transitivy.