How to intuitively think of this definition In our class today instructor said that  
" Let $X$ be ordered set. If any $Y \subset X$ is bounded above, then $X$ is said to have lub property if $\sup(Y)$ exists in $X$."
Lub property states that every non empty subset of real numbers has lub. I cannot relate these two things. Can someone explain this?
Thanks 
 A: Your instructor is wrong. He/she said:

(WRONG) Take any total order $X$, and any $Y⊂X$ that is bounded above in $X$. Then we say that $X$ has the lub property iff $Y$ has a supremum in $X$.

But as Dave said in a comment, the correct definition of the property is:

Take any total order $X$. We say that $X$ has the lub property iff every nonempty $Y⊂X$ that is bounded above in $X$ has a supremum in $X$.
  $\def\eq{\leftrightarrow}$

Notice the swap, and that your instructor missed out the "non-empty" condition! The first is of the form:

$\forall Y ( IsBoundedSubset(Y,X) \to ( LUB(X) \eq HasSupIn(Y,X) ) )$.

Whereas the second is of the form:

$LUB(X) \eq \forall Y ( IsNonemptyBoundedSubset(Y,X) \to HasSupIn(Y,X) )$.

They are not equivalent.
As for what it means, we can see how the correct definition applies to examples.


*

*The integers $\mathbb{Z}$ with the usual ordering has the LUB property because any non-empty set $S$ of integers with an integer $m$ for an upper bound will have a maximum integer (since $\{ m-x : x \in S \}$ is a set of natural numbers and has a minimum).

*The rationals $\mathbb{Q}$ with the usual ordering does not have the LUB property because the set $S = \{ x : x \in \mathbb{Q} \land x^2 < 2 \}$ is non-empty and has an upper bound of $2$ in $\mathbb{Q}$ but no supremum (lowest upper bound) in $\mathbb{Q}$. Why? Suppose there is such a supremum $c$ for $S$ in $\mathbb{Q}$. If $c < \sqrt{2}$, then let $d \in \mathbb{Q}$ such that $c < d < \sqrt{2}$ (by density of $\mathbb{Q}$ in $\mathbb{R}$), and so $d \in S$ contradicting the definition of $c$. Similarly if $c > \sqrt{2}$ then let $d \in \mathbb{Q}$ such that $\sqrt{2} < d < c$, and so $d$ is an upper bound for $S$ in $\mathbb{Q}$, contradicting the definition of $c$. Therefore $c = \sqrt{2}$, which is impossible since $c$ is rational.

*The reals $\mathbb{R}$ with the usual ordering has the LUB property, which is probably the first example you've seen.

*The interval $(0,1)$ with the usual ordering has the LUB property. Why? Take any non-empty $S \subseteq (0,1)$ that is bounded above in $(0,1)$. Let $m \in (0,1)$ be an upper bound for $S$ in $(0,1)$. Let $c = \sup_\mathbb{R}(S)$, which exists because $\mathbb{R}$ has the LUB property. Then $c \le m$ by definition of $c$. Also $c > 0$ because $S$ is non-empty and so has a positive element. Thus $c \in (0,1)$. Also any upper bound for $S$ in $(0,1)$ is also an upper bound for $S$ in $\mathbb{R}$. Hence $c$ is the lowest upper bound for $S$ in $(0,1)$. Therefore, since this holds for any given such $S$, we conclude that $(0,1)$ has the LUB property.
A: It is just a straight generalization of the idea for real numbers to any ordered set:
For $X=\mathbb{R}$ (with the usual ordering)
The least-upper-bound property states that any non-empty $\color{red}{\text{set of real numbers}}$ that has an upper bound must have a least upper bound in $\color{red}{\text{the real numbers}}$.
For any ordered set $X$:
The least-upper-bound property states that any non-empty subset of $\color{red}{X}$ that has an upper bound must have a least upper bound in $\color{red}{X}$.
A: Perhaps an example is helpful.
Take a set $S\subset \mathbb{R}$ to be the image of $f(x)=\sin(x)$.  This set is non-empty (obvious).  This set is bounded above because $\sin(x) < 1000$ (for example).  What the property states is that because this set has an upper bound (1000 in the example) it also has a least-upper-bound.  The least upper-bound in this case is 1 since $\sin(x)\leq 1$.
It may be also helpful to see cases that do not have the least upper-bound property.  For example, if you replace $\mathbb{R}$ with $\mathbb{Q}$ and consider the set $S=\{x\in \mathbb{Q}:x^2\leq 2\}$.  This set also has an upper bound (for example $2$ is an upper bound) but it has no least upper bound (in the rationals).  So $\mathbb{Q}$ does not have the least upper-bound property.  If instead we considered the set in the reals, then it would have a least-upper bound of $\sqrt{2}$.
A: @user21820 : You mis-quoted the professor slightly - he did not say Y had to be bounded above in X, he merely said Y had to be bounded above (full stop).  And the professor was wrong to omit that clause!  Consider your example 4: X = (0,1).  Without that clause one could, for example, take Y = (0,1); then Sup(Y) exists but is obviously not in X, so one would have to conclude that X doesn't have the lub property (despite the fact that it obviously should).
