Example 13, Sec. 3 in Munkres' TOPOLOGY, 2nd ed: Necessary and Sufficient Conditions for a Subset to Have the Least Upper Bound Property

Question

Let $A$ be an ordered set having the least upper bound property, and let $A_0$ be a non-empty subset of $A$.

Then what is (are) the necessary and sufficient condition(s), if any, for $A_0$ to have the least upper bound property under the induced ordered relation?

In the example in Munkres, the subset $(-1, 1)$, for example, of $\mathbb{R}$ does have the least upper bound property, whereas the subset $(-1, 0)\ \cup \ (0, 1)$ does not.

Now we can say with certainty that every (open, closed, bounded, or unbounded) interval in $\mathbb{R}$ does have the least upper bound property. Am I right?

Can we assert anything beyond that for $\mathbb{R}$ for a general ordered set having the least upper bound property?

Answer 1

Incomplete answer.

Let $B\subseteq A_{0}$ be non-empty (with $y\in B$) and bounded above in $A_{0}$ (having $x\in A_{0}$ as upperbound).

Let $z$ denote the least upper bound of $B$ looking at it as a subset of the original $A$.

Then $y\leq z\leq x$.

So if $A_{0}$ has the property $$y,z\in A_{0}\implies\left[y,x\right]\subseteq A_{0}\tag1$$ then the least upperbound of $B$ in $A$ will be an element of $A_{0}$.

Evidently the least upper bound in the original order will also serve as least upperbound in $A_0$.

Apparantly (1) is a sufficient condition for $A_{0}$ to have the least upper bound property. Intervals have this property. I am afraid that the property is even characteristic for intervals. However, there are subsets of $\mathbb R$ that are not an interval and do have the least upper bound property. Example: $(0,1)\cup\{2\}$. So there is no necessity.

Answer 2

I’ll deal just with $\Bbb R$; the general question is more complicated. Suppose that $A\subseteq\Bbb R$ does not have the least upper bound property. Then there are a set $A_0\subseteq A$ and a $b_0\in A$ such that $a\le b_0$ for each $a\in A_0$, but $A_0$ has no least upper bound in $A$. $A_0$ is bounded above by $b$ in $\Bbb R$, so it has a least upper bound $\alpha$ in $\Bbb R$. Clearly $\alpha\notin A$, so $\alpha\notin A_0$, and there is therefore a strictly increasing sequence $\langle a_k:k\in\Bbb N\rangle$ in $A_0$ that converges to $\alpha$ in $\Bbb R$.

Now let $B$ be the set of all upper bounds of $A_0$ in $A$; $b_0\in B$, so $B\ne\varnothing$. Moreover, $a_0<b$ for each $b\in B$, so $B$ is bounded below and therefore has a greatest lower bound $\beta$ in $\Bbb R$. If $\beta\in B$, then $\beta\in A$, and you can prove fairly easily that $\beta$ is the least upper bound of $A_0$ in $A$, so $\beta\notin B$. This means that there is a strictly decreasing sequence $\langle b_k:k\in\Bbb N\rangle$ in $B$ converging to $\beta$ in $\Bbb R$. Clearly we have

$$a_0<a_1<a_2<\ldots<b_2<b_1<b_0\;,$$

where all of these numbers are in $A$. Moreover, for each $k\in\Bbb N$ we have $a_k<\alpha\le\beta<b_k$.

Suppose that $\gamma\in\Bbb R$ with $\alpha\le\gamma\le\beta$. Then $a<\gamma<b$ for each $a\in A_0$ and $b\in B$, so on the one hand $\gamma$ is an upper bound for $A_0$ in $\Bbb R$, and on the other hand $\gamma\notin B$. This is possible only if $\gamma\notin A$, so $[\alpha,\beta]\cap A=\varnothing$. In other words, $A$ contains a strictly increasing convergent sequence $\langle a_k:k\in\Bbb N\rangle$ and a strictly decreasing convergent sequence $\langle b_k:k\in\Bbb N\rangle$ such that if $\alpha=\lim_ka_k$ and $\beta=\lim_kb_k$ in $\Bbb R$, then $\alpha\le\beta$, and $[\alpha,\beta]\cap A=\varnothing$. To put it a little differently, there is no $a\in A$ such that $a_k<a<b_k$ for each $k\in\Bbb N$. Informally we might say that $A$ contains an increasing and a decreasing sequence with nothing lying between them.

Conversely, if $A$ contains such sequences, the set $\{a_k:k\in\Bbb N\}$ is a subset of $A$ that is bounded above in $A$ but has no least upper bound in $A$.

Thus, $A$ has the least upper bound property if and only if it does not contains such a pair of sequences.