# $X \times Y$ 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 $$(M + 1,0) = \sup S$$.

• I have since realised that this counter-example won't work since $[0,1)$ actually is Dedekind complete under the usual order and am working on it. – Ishfaaq Dec 24 '15 at 11:42
• You might want to formulate a few things clearer. A totally oredered set is complete if each non-empty set with an upper bound has a supremum. Hence you should not start with just any subset $S,T\subseteq X$ but rather demand that it be non-empty and have some upper bound. Indeed, if $S=\emptyset$ then $(\pi,\theta)$ may not exist. Also, $X$ need not have a top element, for example $\Bbb R$ is complete but has no top element. - Unless you work with a different definition of "complete" – justanotherhagman Dec 24 '15 at 12:36
• Nah the definition in the notes is different. A totally ordered set is complete if any set has a supremum. And the definition you mention is called Dedekind-Completeness. The notes are here. math.uga.edu/~pete/pointset.pdf – Ishfaaq Dec 24 '15 at 12:40
• Alright. One nitpick then: $\gamma =m$ follows only if $T\ne\emptyset$. To find $\sup T$ for $T=\emptyset$ consider the second component of the supremum of $\emptyset\subseteq X\times Y$ – justanotherhagman Dec 24 '15 at 14:09
• @justanotherhagman: Yes, that's right. Thanks for that. – Ishfaaq Dec 24 '15 at 14:12

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.