In Rudin's book the following are defined as:



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

a) $\alpha$ is an upper bound of $E$

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

Then $\alpha = \sup E$

Least Upper Bound Property


An ordered set $S$ is said to have the LUBP if the following is true:

If $E$ is a proper subset of $S$, $E$ is not empty, and $E$ is bounded above, then $\sup E$ exists in $S$.

Doesn't the first definition imply that if $\sup E$ exists, then $\sup E$ is necessarily in $S$, from the assumption the $\alpha$ is in $S$ ?

And so shouldn't ending of the definition of the LUBP be changed to "$\sup E$ exists.", rather than, "$\sup E$ exists in $S$" ?

  • 1
    $\begingroup$ The proper MathJax representation is $\sup E$, not $supE$. It's coded as \sup E. That not only prevents italicization of $\sup$, but also results in proper spacing to the left and right of $\sup$ in things like $a\sup b$ and in things like $a\sup(b)$ (where there's not as much space to the right of $\sup$), and furthermore when it's in a displayed setting, rather than inline, then it makes subscripts appear directly below $\sup$, as in $\displaystyle a\sup_{b\in B}f(b)\vphantom{\dfrac1{\int}}$ (which is synonymous with $\sup\{f(b) : b\in B\}$). I edited the question accordingly. $\qquad$ $\endgroup$ – Michael Hardy Jan 31 '16 at 16:13
  • $\begingroup$ Thanks. I'll remember that. $\endgroup$ – J.Gudal Jan 31 '16 at 16:40

Let's say we have a set $A \subset \Bbb{Q}$ such that its least upper bound is $\pi$. Clearly, its supremum exists in $S=\Bbb{R}$.

However, its supremum does not exist in $S=\Bbb{Q}$ because we can keep becoming closer and closer to $\pi$ with the rational numbers: $4, 3.2, 3.15, 3.142, 3.1416$, et cetera. Thus, there is no supremum of $A$ in $\Bbb{Q}$ so even if every set in $\Bbb{Q}$ has a supremum that exists (which is true, by the way, since $\Bbb{R}$ has the least upper bound property), $\Bbb{Q}$ does not have the least upper bound property because that supremum might not exist in $\Bbb{Q}$.

Thus, the reason the definition specifies that the supremum must exist in $S$ is because otherwise, the $S$ in the definition of a supremum and the $S$ in the definition of the least upper bound property are not necessarily the same.


Firstly, $E$ does not need to be a proper subset of $S$, just a subset suffices. Consider $$S = E = [0,1]$$ Then $\sup{E}= 1$. Here $1 \in E,S$.

However $\sup{E}$ is not necessarily in $E$ either. To see this consider $S = [0,1]$ and $E = [0,1)$.


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