$X \times Y$ is complete $\implies$ $X, Y$ are complete. Apologies for the long prose. The following problems are from a set of Topology Lecture Notes by Pete L. Clark that I found fascinating and am working through. Am a little unsure about my solutions here. Would be extremely grateful if someone could read through it (again sorry about the length) and give me some feedback.

$X$ and $Y$ are nonempty totally ordered sets. If $X \times Y$,
equipped with the lexicographic order, is complete then so is $X$ and
$Y$.

Proof: The completeness of $X$ is easier to establish and will be attempted first. Let $S$ be a subset of $X$. Since $Y$ is nonempty it contains some point $y$ and $S \times \{y\}$ is a subset of $X \times Y$. Now let $(\pi, \theta) = \sup (S \times \{y\})$.

*

*Let $s \in S$. Then $(s, y) \in S \times \{y\}$. Then, $(s, y) \le (\pi, \theta)$. Then $s \le \pi$.


*Now let $\eta$ be any other upper bound of $S$. Then $(\eta, y)$ is an upper bound of $S \times \{y\}$ which can be easily seen. Hence, $(\pi, \theta) \le (\eta, y)$, so, $\pi \le \eta$.
This shows that $\pi = \sup S$. So $X$ is complete.

The completeness of $Y$ is actually a bigger task. Now let $T$ be an arbitrary subset of $Y$. We have already proved the completeness of $X$. Hence $X$ admits a top element say $m$. Now consider the subset $\{m\} \times T$ of $X \times Y$. This set has a supremum, $(\gamma, \zeta)$. Now it can be easily argued that $\gamma = m$.

*

*Let $t \in T$. Then $(m, t) \in \{m\} \times T \implies (m, t) \le (m, \zeta) \implies t \le \zeta$


*Let $\psi$ be any other upper bound for the set $T$. Then $(m, \psi)$ is an upper bound for the set $\{m\} \times T$. Then, $(m, \zeta) \le (m, \psi) \implies \zeta \le \psi$
So, $\zeta = \sup T$, and therefore $T$ is complete.

Suppose $X \times Y$ is Dedekind complete. What can be said about $X$
and $Y$?

The proof above can be imitated for a nonempty subset $S$ of $X$ to show that $X$ is also Dedekind complete. But the second part of that proof cannot be imitated since there is no reason why $X$ should have a top element.
Counterexample: Let $X = \Bbb N$ and $Y  = [0, 1)$ equipped with the usual orderings. Then we claim that $X \times Y$ is Dedekind complete. Let $S$ be a subset of $X \times Y$ that is bounded above. If $S = \emptyset$ then $(1,0)$, the bottom element of $X \times Y$ serves as the supremum of $S$. So suppose instead that $S$ is nonempty. Then let $S_X, S_Y$ be the projections of $S$ on $X$ and $Y$ respectively. Since $S$ is bounded so is $S_X$ and let $M$ be the top element of $S_X$. If $S_Y$ has a supremum $\lambda$ in $Y$ then $(M, \lambda  ) = \sup S$ if not then $(M + 1,0) = \sup S$.
 A: The completeness arguments are basically just fine, and justanotherhagman has already dealt with the one small nitpick. For Dedekind completeness of $Y$ you can argue as follows.
Suppose that $\varnothing\ne T\subseteq Y$, and $T$ is bounded above in $Y$, say by $y$. Let $x\in X$ be arbitrary, and let $T'=\{x\}\times T$. Then $T'$ is bounded in $X\times Y$ by $\langle x,y\rangle$, so it has a least upper bound $\langle x',y'\rangle$ in $X\times Y$. Clearly $\langle x',y'\rangle\le\langle x,y\rangle$, so $x'\le x$. On the other hand, $T\ne\varnothing$, so there is some $\langle x,t_0\rangle\in T'$, and since $\langle x,t_0\rangle\le\langle x',y'\rangle$, we must have $x\le x'$ and hence $x'=x$. It follows that $\langle x,t\rangle\le\langle x,y'\rangle$ for each $t\in T$ and hence that $t\le y'$ in $Y$ for each $t\in T$, i.e., that $y'$ is an upper bound for $T$ in $Y$. I’ll leave it to you to finish off the argument by showing that $y'$ actually the least upper bound of $T$.
Alternatively, you could suppose that $T$ is a non-empty subset of $Y$ that has an upper bound but no least upper bound, define $T'$ as above, and show that in $X\times Y$ the set $T'$ is bounded above but has no least upper bound; the argument is essentially the same.
