I'm trying to read through and understand this proof from Rudin, but am a bit confused at just a few steps. I'm going to try to replicate the proof and pause at those particular steps.

Theorem. Suppose $S$ is an ordered set with the least-upper-bound property, $B \subset S$, $B$ is not empty, and $B$ is bounded below. Let $L$ be the set of all lower bounds of $B$. Then $\alpha = \sup L$ exists in $S$, and $\alpha = \inf B$.

Proof. Let $B \subset S$, where $S$ is an ordered set and $B$ is non-empty and bounded below. Define $L$ to the set of all lower bounds of $B$, which is non-empty due to the fact that $B$ is bounded below.

(1) This is where my first question comes in: Rudin argues here that $L$ consists of all elements $y \in S$ s.t. $y \leq x$ for every element $x \in B$. I understand why this holds for all elements of $B$, as this is the definition of lower bounds. But, we aren't given the size or nature of $S$ (finite, infinite, etc.). If $S\neq \mathbb{R}$, would it noe be possible for us to have other element not in $S$ which bounds $B$ below? We also don't know whether $S$ is bounded below; if so, there exist lower bounds of $B$, and thus elements of $L$, which are in $\mathbb{R}$ but not $L$. So, might it be more accurate to say that $L = \{y \in \mathbb{R} : y \leq x, \forall x \in B\}$?

Continuing the proof, and assuming Rudin's last step as to the definition of $L$, it's clear that every $x$ is an upper bound of $L$, by definition, and thus every element of $B$ bounds $L$ above.

(2) Here, Rudin argues that the properties we assumed about $S$ tell us that $L$ has a supremum, $\alpha$, in $S$. Here, I am not quite following. Clearly, $S$ is ordered and has a least-upper bound. We don't know whether $B$ contains that least upper bound, as we don't know the nature of the subset. Does this step require -- or perhaps embed, without proof -- a lemma that a subset of a ordered set with a least upper bound also has a least upper bound, which is less than or equal to the least upper bound of the one contained in its superset? I'm also struggling a bit to see why this element must exist in $S$, but this may be that not only is every element of $B$ an upper bound of $L$, but that $B$ contains all such upper bounds of $L$. Is this correct? Would this not presuppose that $B$, and thus $L$ (as the largest element of $L$ and the smallest element of $B$ are the same), are composed of real numbers, and that this list is "complete"? Otherwise, we can find a real that is in neither set but bounds $B$ below.

So, again assuming this last step, we define the supremum of $L$ as some element $\alpha \in S$. We then consider some arbitrary element $\gamma$, which has the property that $\gamma < \alpha$. (Rudin does not state that this element is in $S$, but I'm not sure that it quite matters here.) Since $\gamma$ is the least-upper-bound of $L$, $\alpha$ cannot be an upper bound of $L$. But, all elements of $B$ are upper bounds of $L$, so $\gamma \not \in B$. Hence, $B$ cannot contain any element less than $\alpha$, and we have $\alpha \leq x, \forall x \in B$.

(3) I understand everything from the above except, perhaps, the final line that I have for the moment omitted. Rudin concludes from this that $\alpha \in L$. Is the argument here, again, that because $L$ contains all elements $y \leq x$ for $x \in B$, that $\alpha$ is simultaneously in line and $B$ due to the fact that we could have $y = x$? (In other words, assuming some notion of 'completeness,' the largest element of $L$ is simultaneously the smallest element of $B$? I hope I'm not assuming the conclusion of the proof here.)

So, again assuming the above that $\alpha \in L$, we consider some element $\beta$. (Rudin again makes no assumptions on whether $\beta$ lives in $S$, so I assume we don't need to make any assumptions, and can just take $\beta$ to be a real number.) Then, we let $\beta > \alpha$.

(4) From here, I am having difficulty seeing how Rudin concludes that $\beta \not \in L$. It seems that the argument is that, since $\alpha$ is the least upper bound of $S$, and every element of $L$ (per Rudin, though I questioned why this is the case above) is also in $S$, that clearly $\alpha$ must also bound this subset of $S$ above. So, clearly, a greater element, $\beta$, can't be in this set that's bounded above by $\alpha$. Is this correct? If so, this suggests that the argument that $L$ contains elements only in $S$ is even more crucial, though I still am struggling to see why this is the case.

From here, the remainder of the proof is easy: $\alpha \in L$, as we showed above, but any greater element than $\alpha$, such as $\beta$ from above, cannot be in $L$, so $\alpha$ is not only an upper bound of $L$, but the greatest such upper bound. Since $L$ contains all of the lower bounds of $B$, this implies that $\alpha$ is the greatest lower bound, so $\alpha = \inf B$.

I'd appreciate any help and insights on the questions I raised above. Hopefully I have understood and correctly framed at least parts of this proof.

Thanks in advance.


2 Answers 2

  1. What has $\mathbb R$ to do with this? We are talking about an arbitrary ordered set, which may well have nothing to do with $\mathbb R$.
  2. The set $S$ has the least upper bound property and $L$ has un upper bound (every element of $B$ is an upper bound of $L$). Therefore, $\sup L$ exists.
  3. $\alpha\in L$ because by definition $L$ is the set of all lower bounds of $B$ amd because Rudin has just proved that $\alpha$ is a lower bound of $B$.
  4. If $\beta>\alpha$, then $\beta\notin L$ because $\alpha=\sup L$.

In your question, you mention the possibility that some element doesn't belong to $S$. That's not possible. In this context, $S$ is the whole universe. There is nothing outside of it.

  • $\begingroup$ Thank you for this. I didn't think to consider $S$ the universe, but this really seems to make the entire proof make sense to me. $\endgroup$
    – user465188
    Commented Aug 30, 2018 at 14:34
  • $\begingroup$ @Matt.P I'm glad I could help. $\endgroup$ Commented Aug 30, 2018 at 14:35

I found the proof slightly confusing too. Jose's answer is correct, but slightly terse, so I figured I would flesh out some of the steps to the proof that weren't apparent to me on first reading, just in case anyone else reads the proof and stumbles across this thread looking for guidance.

For (1.) Note that definition 1.7 states that for a subset $E$ to be bounded above, that means that there must exist some $\beta \in S$ such that $\forall x \in E$, we have that $x \leq \beta$. The definition of "bounded below" is similar. So $L$ must be nonempty by the textbook's provided definition of what it means for $E$ to be bounded below.

For (2.) The reason why we know that $S$ contains the supremum of $L$ is that we are given that $S$ has the least upper bound property. Given that $S$ satisfies this property, and given that $L$ is a nonempty subset of $S$ that is bounded above (because every element of $B$ is an upper bound of $L$ and $B$ is non-empty), we can conclude that $\alpha = $ sup $L \in S$

For (3.) we want to prove that $\alpha \in L$. Consider, by way of contradiction, some arbitrary element $y < \alpha$ where $y > x$ for some $x \in B$. Every element of $B$ is an upper bound of $L$ so this would imply that $L$ has an upper bound that is less than $\alpha$. But this contradicts our assumption that $\alpha$ was the least upper bound of $L$. Hence, for every element $y$ such that $y < \alpha$, we can conclude that $\forall x \in B, y \leq x$. Similarly, if we take some arbitrary element $x \in B$, then we have just shown that it is not possible that $x < \alpha$. Hence $\alpha$ is a lower bound of $B$. But if $\alpha$ is a lower bound of $B$, that implies that $\alpha \in L$, because $L$ is defined as the set of all lower bounds of $B$.

(4.) If $\beta > \alpha$, then it cannot be the case that $\beta \in L$, because we assumed that $\alpha$ was the least upper bound of $L$, and $\alpha$ would not be a true upper bound of $L$ if there were an element of $L$ greater than $\alpha$.

We have shown that $\alpha$ is a lower bound of $B$ (because we proved that $\alpha \in L$, the set of all lower bounds of $B$. And moreover, we have shown that if some element $\beta$ is greater than $\alpha$, then it is not in $L$, because $\alpha$ is the least upper bound of $L$ and it would not be an upper bound if there were an element greater than it that belonged to $L$. Hence, there is no element $\beta$ larger than $\alpha$ that is a lower bound of $B$. Hence, $\alpha$ is the largest lower bound of $B$. Thus $\alpha$ = inf $B$ exists in $S$.


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