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Here is Prob. 1, Sec. 27, in the book Topology by James R. Munkres, 2nd edition:

Prove that if $X$ is an ordered set in which every closed interval is compact, then $X$ has the least upper bound property.

My Attempt:

Let $A$ be any non-empty subset of $X$ such that $A$ is bounded above in $X$; let $U$ be the set of all the upper bounds in $X$ of this set $A$. Then $U$ is also a non-empty subset of $X$.

Case 1.

If $A \cap U \not= \emptyset$, let us suppose that $v \in A \cap U$.

Now as $v \in U$, so $v$ is an upper bound of the set $A$, by our definition of the set $U$.

Also, as every element of $U$ is an upper bound of the set $A$ and as $v \in A$, so we must have $$ v \leq u \ \mbox{ for every element } u \in U. $$

Thus $v$ is an upper bound of the set $A$ and $v \leq u$ for every upper bound $u$ of $A$. Therefore $v$ is the least upper bound of the set $A$.

Case 2.

Let us now suppose that $A \cap U = \emptyset$.

Then for every element $a \in A$ and for every element $u \in U$, we must have $a < u$.

Let $a \in A$ and $u \in U$ be arbitrary. Then $a < u$. Moreover, the closed interval $[a,u]$ is compact, by our hypothesis.

Let $\mathscr{L}$ be the collection of all the closed intervals of the form $[x, y]$, where $x \in A$, $y \in U$, and $[x,y] \subset [a,u]$. That is, let $$ \mathscr{L} \colon= \big\{ \ [x, y] \ \colon \ x \in A, y \in U, a \leq x < y \leq u \ \big\}. \tag{*} $$ Then the interval $[a, u]$ itself is in $\mathscr{L}$ so that $\mathscr{L}$ is non-empty.

If $\left[x_1, y_1 \right], \ldots, \left[x_n, y_n \right]$ be any finite subcollection of $\mathscr{L}$, then we have $$ \bigcap_{j=1}^n \left[ x_j, y_j \right] = \left[ x_0, y_0 \right], $$ where $$ x_0 \colon= \max \left\{ x_1, \ldots, x_n \right\}, \qquad \mbox{ and } \qquad y_0 \colon= \min \left\{ y_1, \ldots, y_n \right\}. $$ Thus $\bigcap_{j=1}^n \left[ x_j, y_j \right]$ is again an interval in $\mathscr{L}$ and is therefore non-empty.

Then $\mathscr{L}$ is a collection of non-empty closed sets in $[a,u]$ such that $\mathscr{L}$ has the finite intersection property. Since $[a,u]$ is compact, therefore the intersection of all the closed intervals in $\mathscr{L}$ is non-empty, by Theorem 26.9 in Munkres.

That is, there is an element $p \in [a,u]$ such that $p \in [x,y]$ for all $x \in A$ and for all $y \in U$ for which $[x,y] \subset [a,u]$. In particular, this $p \in [a, u]$ also, so that $$ a \leq p \leq u. \tag{1} $$

But $a$ was chosen to be an arbitrary element of set $A$, and $u$ was chosen to be an arbitrary element of the set $U$ (i.e. $u$ was chosen to be an arbitrary upper bound for the set $A$).

Thus the leftmost inequality in (1) implies that $p$ is an upper bound of the set $A$, and the rightmost inequality in (1) implies that $p$ is also the least of all of the upper bounds of $A$.

Hence $p$ is the least upper bound of the set $A$.

But as $A$ was chosen to be an arbitrary non-empty subset of $X$ which was bounded above in $X$, so we can conclude that $X$ has the least upper bound property.

Is this proof correct? If so, is it clear enough? If not, then where are the issues?

Cameron Buie
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Saaqib Mahmood
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    Perhaps add a detail about how there is only one point $p$ and not a whole interval $[p_0,p_1]$. It's not very important, though. – Arthur May 21 '15 at 21:20
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    @Arthur, please have a look at my post now; I've editted it to add the uniqueness part. How do you like the proof now? – Saaqib Mahmood May 21 '15 at 21:38
  • You have to show $p \in U$ before you can conclude $[a, p] \in \mathscr L$, but that's just a matter of which order you write things. Otherwise it looks fine to me. – Arthur May 22 '15 at 05:28
  • I am having a tough time showing that $\mathscr{L}$ has the finite intersection property. Could someone show me this? – user193319 Jun 10 '17 at 17:49
  • Also, why do we have $p \in [x,y]$ for all $x \in A$ and $y \in U$. How do you know that every $x$ is in $A$ and every $y$ is in $U$? There doesn't seem to be much justification for many of the steps in the proof. – user193319 Jun 10 '17 at 17:54
  • @Arthur can you please review my post now? I'm sure you'll find it better this time! – Saaqib Mahmood Oct 02 '18 at 09:30
  • @user193319 please have a look at my post now. I've improved the proof. I'm sure you'll understand it now! – Saaqib Mahmood Oct 02 '18 at 09:31
  • @SaaqibMahmood your proof seems correct and very clear, to me. – João Alves Jr. Mar 29 '21 at 21:09
  • How to show $\bigcap_{[x,y] \in \mathscr{L}} [x,y] ={ p}$? and $p$ is independent of $a$ and $u$, i.e. if $a\neq a’ \in A$ and $u\neq u’\in U$, then $\bigcap \mathscr{L}’ ={p}$, where $\mathscr{L}’ ={ [e,f]| e\in A, f\in U $ and $[e,f]\subseteq [a’,u’]}$? – user264745 Mar 31 '22 at 13:58
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    @user264745, it isn't particularly necessary to show that $\bigcap\mathscr{L}={p},$ as it turns out, but it is true, and follows readily from proving that $p$ is the least element of $U.$ As far as independence from our choice of $a\in A$ and $u\in U,$ see point 9 in my proof outline below. – Cameron Buie May 03 '24 at 22:47
  • @JoãoAlvesJr. It was very close! It went off the rails in the third-to-last paragraph ("Thus the leftmost inequality...."). See my answer below for analysis. – Cameron Buie Jul 31 '24 at 14:04

1 Answers1

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To cut down on the number of unanswered questions on the site, here is a (very belated) answer to your question.

First, I will go through and offer some critiques to specific parts of your proof, then I'll make a few other suggestions.

Your first case looks fine, but your second case has some issues.

Let $\mathscr{L}$ be the collection of all the closed intervals of the form $[x, y]$, where $x \in A$, $y \in U$, and $[x,y] \subset [a,u]$. That is, let $$ \mathscr{L} \colon= \big\{ \ [x, y] \ \colon \ x \in A, y \in U, a \leq x < y \leq u \ \big\}. \tag{*} $$ Then the interval $[a, u]$ itself is in $\mathscr{L}$ so that $\mathscr{L}$ is non-empty.

Everything is okay up to and including the part above. I would note at this point that, since for any $x\in A$ and any $y\in U$ we have $x<y,$ it follows that $x$ and $y$ are elements of $[x,y],$ and so $\mathscr{L}$ is therefore a set whose elements are also non-empty sets. You take this fact for granted later without proof.

If $\left[x_1, y_1 \right], \ldots, \left[x_n, y_n \right]$ be any finite subcollection of $\mathscr{L}$, then we have $$ \bigcap_{j=1}^n \left[ x_j, y_j \right] = \left[ x_0, y_0 \right], $$ where $$ x_0 \colon= \max \left\{ x_1, \ldots, x_n \right\}, \qquad \mbox{ and } \qquad y_0 \colon= \min \left\{ y_1, \ldots, y_n \right\}. $$ Thus $\bigcap_{j=1}^n \left[ x_j, y_j \right]$ is again an interval in $\mathscr{L}$ and is therefore non-empty.

It isn't actually important that $\bigcap_{j=1}^n[x_j,y_j]\in\mathscr{L},$ only that $\bigcap_{j=1}^n[x_j,y_j]$ is non-empty. To prove that, it is sufficient to show that $x_0\in\bigcap_{j=1}^n[x_j,y_j]$ or that $y_0\in\bigcap_{j=1}^n[x_j,y_j],$ either of which can be proved fairly easily.

If you really want, you can prove $\bigcap_{j=1}^n[x_j,y_j]=[x_0,y_0],$ such as by using double-inclusion. The interval $[x_0,y_0]$ will be non-empty if and only if $x_0\leq y_0,$ the latter of which follows readily from the fact that $x_0\in A$ and $y_0\in U.$

Then $\mathscr{L}$ is a collection of non-empty closed sets in $[a,u]$ such that $\mathscr{L}$ has the finite intersection property. Since $[a,u]$ is compact, therefore the intersection of all the closed intervals in $\mathscr{L}$ is non-empty, by Theorem 26.9 in Munkres.

That is, there is an element $p \in [a,u]$ such that $p \in [x,y]$ for all $x \in A$ and for all $y \in U$ for which $[x,y] \subset [a,u]$. In particular, this $p \in [a, u]$ also, so that $$ a \leq p \leq u. \tag{1} $$

Here, with $(1),$ you've ignored the vast majority of what you actually proved, and it will shortly cause a major problem.

But $a$ was chosen to be an arbitrary element of set $A$, and $u$ was chosen to be an arbitrary element of the set $U$ (i.e. $u$ was chosen to be an arbitrary upper bound for the set $A$).

Thus the leftmost inequality in (1) implies that $p$ is an upper bound of the set $A$, and the rightmost inequality in (1) implies that $p$ is also the least of all of the upper bounds of $A$.

It is here that your proof fails. You are correct that $p$ is the least upper bound of $A,$ but it does not follow from the premise $$\forall x\in A,\forall y\in U,\exists p\in X:x\leq p\leq y.\tag{$1'$}$$

Consider $X=\mathbb{Q},$ $A:=\{p\in X:p<\pi\},$ and (thus) $U=\{p\in X:p>\pi\}.$ Then $(1')$ holds, but no such $p$ will be the least upper bound of $A,$ and if we were to define $\mathscr{L}$ as before, then it would have the finite intersection property and all of its elements would be non-empty (in fact, infinite) sets, but we would have $\bigcap\mathscr{L}=\emptyset.$

Instead, you should be using the non-emptiness of $\bigcap\mathscr{L}$ (along with the hypothesis that $X$ is totally ordered) to conclude that the following holds: $$\exists p\in X:\forall x\in A,\forall y\in U,x\leq p\leq y.\tag{$1''$}$$

Aside from the adjustments I mentioned above, I would probably have structured the proof a bit differently in terms of order.

Here's an outline of how I would go about it.

  1. Suppose that $X$ is an ordered set with the property that every closed and bounded interval is compact. (This is a weaker assumption a priori, but turns out to be sufficient.) Take an arbitrary non-empty subset $A$ of $X$ that is bounded above in $X,$ and define $U$ as you did.
  2. Observe that, since $A$ is an arbitrary non-empty subset of $X$ that is bounded above in $X,$ then in order to show $X$ has the least upper bound property, it suffices to show that $U$ necessarily has a least element.
  3. Break things up into two cases as you did.
  4. Handle Case 1 as you did.
  5. Show that in Case 2, for any $x\in A$ and any $y\in U,$ we have that $x<y,$ and so $[x,y]$ is non-empty.
  6. Take an arbitrary $a\in A$ and an arbitrary $u\in U,$ observe that $[a,u]$ is a closed and bounded interval in $X,$ so is compact by hypothesis, and then define $\mathscr{L}$ as you did.
  7. Prove that $\mathscr{L}$ is non-empty and has the finite intersection property, basically as you did, with the slight adjustments I mentioned above.
  8. Conclude from compactness of $[a,u]$ that there is some $p\in\bigcap\mathscr{L}.$
  9. Prove that (almost by definition of $\mathscr{L}$) for any $x\in A$ with $a\leq x,$ we have $x\leq p;$ then, use the fact that $X$ is totally ordered to prove that for any $x\in A,$ we have $x\leq p.$
  10. Note that we can similarly prove that for any $y\in U,$ we have $p\leq y.$
  11. Conclude that $p$ is the least element of $U,$ thus completing the proof.
Cameron Buie
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