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?


  • 4
    $\begingroup$ If that's what your instructor literally said, I don't see how anyone who doesn't already know what is intended could make sense of it. For example, if the antecedent of the main conditional is false (we only need the existence of an unbounded subset), then I suppose $X$ automatically has the lub property?!? Try this: Let $X$ be an ordered set. We say that $X$ has the least upper bound property if the following property holds: For each $Y,$ if $Y \neq \emptyset$ and $Y \subseteq X$ and $Y$ is bounded above in $X,$ then $\sup (Y)$ exists. $\endgroup$ Commented Jul 29, 2016 at 14:08
  • $\begingroup$ @DaveL.Renfro: Your comment should have been an answer! Anyway I've written one to make the logical structure absolutely clear. =) $\endgroup$
    – user21820
    Commented Jul 29, 2016 at 14:31
  • $\begingroup$ I think the confusion here surrounds the word "any" and the arrangement of the conditionals and the quantifiers. I think the intended phrasing was "If $X$ is such that for any $Y \subset X$ which is bounded above, $\sup(Y)$ exists in $X$, then $X$ has lub property". This is clearer because it puts all the antecedents before the conclusion, and it puts the quantifier in a clearer position. $\endgroup$
    – Ian
    Commented Jul 29, 2016 at 14:58
  • $\begingroup$ @Ian: You're simply re-interpreting the instructor to make it correct. One can re-interpret any writing to make it correct. Furthermore, it's still wrong... See my answer for what else is wrong. $\endgroup$
    – user21820
    Commented Jul 29, 2016 at 15:05

4 Answers 4


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.

  1. 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).

  2. 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.

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

  4. 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.

  • $\begingroup$ I'm not sure your reading is quite correct. Sometimes "any" means a universal, sometimes it means an existential, we must be careful about the context. Certainly the intent here (as we know from more carefully worded definitions of the same term) was a universal. But you've read it as an existential. $\endgroup$
    – Ian
    Commented Jul 29, 2016 at 15:00
  • $\begingroup$ @Ian: What are you talking about?? I gave the correct definition of the LUB property. The one given by the instructor is wrong. If you insist on reading the "any" in the instructor's version as "some", then it becomes completely meaningless, even ignoring the problem with the swapping. $\endgroup$
    – user21820
    Commented Jul 29, 2016 at 15:04
  • 1
    $\begingroup$ Your definition is of course correct. The definition in the OP is poorly worded. I am saying that I think you are reading it differently from how it was intended so that you can say that it is wrong rather than merely being poorly worded. I'm saying that you're doing that by reading the "any" as an existential when it should of course be a universal. If "any" were replaced by "some" then of course it would be objectively wrong, but it isn't. It is just very badly worded, with the primary problem being the dangling conditional on the end, which should be inside the antecedent. $\endgroup$
    – Ian
    Commented Jul 29, 2016 at 15:09
  • $\begingroup$ @Ian: Oh I didn't read it as an existential by the way. My rendering of the instructor's wordings was "Take any ... and any ..." which is a universal. I also made clear in the symbolic form that the instructor had the universal quantifier before the equivalence. $\endgroup$
    – user21820
    Commented Jul 29, 2016 at 15:34
  • $\begingroup$ You're technically right, your reading is apparently not an existential (though I did wind up reading it that way, partly because of "take"). Nonetheless it appears to behave much like an existential. Indeed, in your version of the instructor's statement, you say that for any $Y$, if $Y$ is bounded in $X$, then the truth value of LUB(X) is the same as the truth value of HasSupIn(Y,X). But the truth value of LUB(X) does not involve $Y$. So as soon as you have one $Y$ bounded in $X$ with sup in $X$, LUB(X) holds. $\endgroup$
    – Ian
    Commented Jul 29, 2016 at 16:05

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 $\color{red}{X}$.

  • 1
    $\begingroup$ +1 I think you figured out what the OP was really asking. $\endgroup$ Commented Jul 29, 2016 at 13:21
  • $\begingroup$ Maybe it helps to be even more explicit: let $\mathcal Y$ be the collection of subsets of $X$ (endowed with an order) that are bounded from above. I.e. $Y\in\mathcal Y$ then $Y\subset X$ and has an upper bound. Let $S$ be the set of suprema of the sets in $\mathcal Y$. I.e. $s\in S$ then $s=\sup Y$ for some $Y\in\mathcal Y$. We say that $X$ has the lub property if $S\subset X$. I.e. all suprema must be members of $X$, just so the point had no chance of being overlooked. $\endgroup$
    – jdods
    Commented Jul 29, 2016 at 13:34
  • $\begingroup$ @jdods My problem with that is that it assumes some ambient completed order for $S$ to live in. If we haven't constructed $\mathbb{R}$ yet then it is not so clear what this completed order should be for $\mathbb{Q}$. $\endgroup$
    – Ian
    Commented Jul 29, 2016 at 15:03

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 $\sqrt{2}$.

  • $\begingroup$ It might be good to consider those numbers that square to less than 2 so you don't have to go outside of the rationale when defining your interesting set, and instead have something that could be considered in any ordered field. $\endgroup$
    – Mark S.
    Commented Jul 29, 2016 at 13:37
  • $\begingroup$ Also its more important that it has an upper bound that's in the rationals, since otherwise it wouldn't show the rationals don't have the least upper bound property. $\endgroup$
    – Mark S.
    Commented Jul 29, 2016 at 13:41
  • $\begingroup$ @MarkS., another good suggestion. Thanks. I made the edit. $\endgroup$
    – TravisJ
    Commented Jul 29, 2016 at 13:43
  • $\begingroup$ Might be interesting to mention that least upper bound property is pretty much how the reals are defined. Natural numbers arise from having a "next" operator; you can get integers, rationals and reals by defining addition, multiplication and polynomials and requiring that every equation has a solution (IOW, declaring that every expanding your set of numbers by declaring that every equation represents a number). You then get from algebraic to real by adding all the missing upper bounds (declaring that every bounded set represents a number). $\endgroup$
    – Tgr
    Commented Jul 29, 2016 at 16:49

@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).

  • $\begingroup$ I was lenient about that because often when we talk about a subset of a total order and say "bounded above" we mean (by the context) "bounded above in that total order". The reason I was lenient is that "bounded above [full-stop]" is really meaningless because without any ambient ordering there's no meaning to boundedness. Note that your example has $(0,1)$ being bounded above in $\mathbb{R}$, which is the default interpretation because that's the ambient ordering on reals. $\endgroup$
    – user21820
    Commented Jul 29, 2016 at 16:08

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