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Definition: Let $A$ the set with order relation. We say that the set $A$ has least upper bound property if any $A_0\subset A$, $A_0\neq \varnothing$ which has upper bound has the least upper bound.

Question 1: When we say "has upper bound..." do we mean that its upper bound is in $A$?

Question 2: When we say "has the least upper bound..." do we mean that its least upper bound is in $A$?

Example: Consider the set $A=(-1,1)$ of real numbers in the usual order. Assuming the fact that the real numbers have least upper bound property, it follows that the set $A$ has the least upper bound property (why?). For given any subset of $A$ having an upper bound in $A$ , it follows that its least upper bound must be in $A$. For example, the subset $\{-1/2n: n\in \mathbb{N}\}$ of $A$, thought it has no largest element, does have a least upper bound in $A$, the number $0$.

$ \quad $ On the other hand, the set $B=(-1,0)\cup (0,1)$ does not have th least upper bound property . The subset $\{-1/2n: n\in > \mathbb{N}\}$ of $B$ is bounded above by any element of $(0,1)$, but it has no least upper bound in $B$.

I have read this example very carefully and I guess that it provides an example of subsets of reals which has LUB-property and has not, respectively.

Do I correctly interpreted the meaning of above example?

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    $\begingroup$ Q1&2. Yes.${}{}$ $\endgroup$ Commented Sep 2, 2018 at 23:24

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Your point is that if a set $A$ has the least upper bound property, it does not imply that every subset of A also has the least upper bound property.

Yes, you are quite right.

A good example is the set of rational numbers which does not have the least upper bound property while it is a subset of real numbers which has the least upper bound property.

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