showing chain-completeness or conditional completeness of functions Q: Given a nonempty set $X$ and a poset ($Y$, $\succcurlyeq$),
    define the partial order $\trianglerighteq$ on $Y^X$ by $f
\trianglerighteq    g$ if and only if $f(x) \succcurlyeq g(x)$ for each $x\in X$($Y^X$
    stands for the collection    of all functions from $X$ to $Y$). Show
    that if ($Y$, $\succcurlyeq$) is chain-complete(or conditionally
    complete), so is ($Y^X$,    $\trianglerighteq$).
I attempted to prove this claim as follows.


*

*(Chain-completeness)Let $h_1, h_2, ..$ be functions from $X$ to $Y$ 
such that they form chains. For every chain in subsets of $Y^X$,
there exists a supremum if there exists a supremum of $y$ for any $x
\in X$.

*(Conditionally completeness) For every subset $S \subseteq Y^X$ we have a supremum if every subset of $Y$ has an $\succcurlyeq$-upper bound and a supremum.


I feel like this is a regurgitation of concepts rather than a rigorous proof. Any suggestions for formalizing my ideas are welcome.
 A: I would be a little more explicit.

*

*Given a chain $\langle h_i\mid i\in I\rangle\subset X^Y$, define a function $h:X\to Y$ as follows: for any $x\in X$, let $h(x)$ be the supremum of the chain $\langle h_i(x)\mid i\in I\rangle$. This supremum exists, since $Y$ is chain-complete. Then $h$ is the supremum of $\langle h_i\mid i\in I\rangle\subset X^Y$.
To show this, you prove that $h$ is an upper bound, i.e. $h_i\trianglelefteq h$ for each $i\in I$, and that $h$ is minimal, i.e. for any $f$ such that $h_i\trianglelefteq f$ for all $i\in I$ we also have $h\trianglelefteq f$.


*Suppose $S\subset Y^X$ and there exists some upper bound $f$ such that $s\trianglelefteq f$ for all $s\in S$, then we have to show there exists an $h:X\to Y$ such that $h$ is the supremum of $S$. We define this $h$ by noting that for each $x\in X$ the set $\{ s(x)\mid s\in S\}\subset Y$ has $f(x)$ as upper bound, and thus the supremum $a$ of  $\{s(x)\mid s\in S\}$ exists. We then define $h(x)=a$.
Again, to be completely rigorous, you have to show that this $h$ is indeed the supremum.
