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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 $(M + 1,0) = \sup S$.

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  • $\begingroup$ 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. $\endgroup$ – Ishfaaq Dec 24 '15 at 11:42
  • $\begingroup$ 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" $\endgroup$ – justanotherhagman Dec 24 '15 at 12:36
  • $\begingroup$ 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 $\endgroup$ – Ishfaaq Dec 24 '15 at 12:40
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    $\begingroup$ 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 $\emptyset\subseteq X\times Y$ $\endgroup$ – justanotherhagman Dec 24 '15 at 14:09
  • $\begingroup$ @justanotherhagman: Yes, that's right. Thanks for that. $\endgroup$ – Ishfaaq Dec 24 '15 at 14:12
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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.

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