I am reading Rudin's Principles of Mathematical Analysis in order to prepare for my first course in Real Analysis I intend to take this fall. The book just defined what an upper bound is and then defined supremum/ least upper bound as:

Suppose $S$ is an ordered set, with $E$ as a subset of $S$, where $E$ is bounded above. Suppose there exists an $\alpha \in S$ with the following properties:

A) $\alpha$ is an upper bound of $E$.

B) If $\gamma < \alpha$ then $\gamma$ is not an upper bound of $E$.

I do not understand the difference between upper bound and least upper bound. If someone could explain the difference between the two and possibly provide an example, it would be much appreciated. Thanks.

  • $\begingroup$ Another way to phrase B) would be "If $\gamma$ is an upper bound of $E$ then $\gamma \geq \alpha$." $\endgroup$
    – name
    Aug 12 '13 at 20:32
  • $\begingroup$ This book is using a negative approach to defining the least upper bound; it is choosing a number outside the possibility of being an upper bound as defining the limit of what the least upper bound can be. The answer below has demonstrated a positive example of a comparison upper bound. The positive approach would be to say that every upper bound is greater than or equal to the least upper bound. $\endgroup$
    – abiessu
    Aug 12 '13 at 20:33
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    $\begingroup$ An upper bound for a set S of numbers is an element $\alpha $ that is larger than every element in S. Notice that if $\alpha$ is an upper bound, then any element $\beta$ larger than $\alpha $ is also an upper bound, as if $\beta > \alpha$, then $\beta$ is also larger than any element in S. But, out of all the upper bounds for a set S, there is a least element that is also an upper bound. For, say, the set of negative reals, any positive number is an upper bound, but the number $0$ is the least number that is larger than all negative numbers. $\endgroup$
    – FBD
    Aug 12 '13 at 20:38

Every least upper bound is an upper bound, however the least upper bound is the smallest number that is still an upper bound. Example: Take the set $(0,1)$. It has $2$ as an upper bound but clearly the smallest upper bound that the set can have is the number $1$ and hence it's the least upper bound.

  • $\begingroup$ Thanks for your help and example. I understand now. $\endgroup$ Aug 13 '13 at 0:39
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    $\begingroup$ To whomever approved the edit on my answer: please actually read the answer before approving things. That was a completely erroneous approval. $\endgroup$ Jan 18 '17 at 15:55
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    $\begingroup$ @CameronWilliams I'm trying to understand the concept of $lup$ and $glb$. So if you meant the interval $(0,1)$ then $1$ is the $lub$ of that interval as $0.9999... =1$ ? $\endgroup$ Jan 18 '17 at 16:20
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    $\begingroup$ @Michael that is correct! $\endgroup$ Jan 18 '17 at 20:17

Maybe you like this definition better?

Let $A$ be nonempty. We say that $\alpha$ is the least upper bound of $A$ if

$(1)$ It is an upper bound of $A$, that is, if $x\in A$; then $x\leq \alpha$.

$(2)$ If $\beta$ is any other upper bound $\alpha\leq \beta$. That is, $\alpha$ is the least of all upper bounds of $A$.

As you can see the l.u.b. has the unique property $(2)$. Why unique? Because if $\gamma$ is another l.u.b., by definition, we must have both $\alpha\leq \gamma$ and $\gamma\leq \alpha$, but this means we must have $\alpha=\gamma$. So l.u.b.s when they exist, are unique.

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    $\begingroup$ The treatment at the last part was magnificent.+1. $\endgroup$
    – user142971
    May 17 '15 at 15:15

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