# Question on Theorem 1.11 in Baby Rudin

Definition 1.7 in Baby Rudin says: Suppose S is an ordered set, and E $$\subset$$ S. If there exists a $$\beta \in$$ S such that $$x \leq \beta$$ for every $$x \in E$$, we say E is bounded above and call $$\beta$$ an upper bound of E.

In the proof of Theorem 1.11 in Baby Rudin, Rudin says that all x in B are upper bounds of L, thus L is bounded above. So Rudin is saying $$\forall x \in B$$ $$\forall l \in L$$, $$l \leq x$$, thus L is bounded above. How can we say all x in B are upper bounds of L if the definition uses there exists? $$\forall x \in B$$ $$\forall l \in L$$, $$l \leq x.$$ doesn't match definition 1.7. If B were the empty set $$\forall x \in B$$ $$\forall l \in L$$, $$l \leq x.$$ would be vacously true but saying all x in the empty set are upper bounds of L doesn't make sense to me because there are no x in the empty set. Also how would those x's in the empty set be elements of S? Can someone explain this please? Thank you!

• It might be helpful to state the context of theorem $1.11$ (and the rest of the proof). Commented Jun 17 at 18:30
• Sounds good I'll add that! Commented Jun 17 at 18:32

## 2 Answers

It seems to me you are confused about what exactly Definition 1.7 is defining. It is actually defining two things: what it means for a set to be bounded above, and what it means for an element to be an upper bound.

Maybe it’ll be easier for you to understand like this. Let us define “upper bound”:

Suppose $$S$$ is an ordered set and $$E\subset S$$. We say $$\beta$$ is an upper bound of $$E$$ iff:

$$\forall x\in E, \beta \geq x$$

Now let us define “bounded above”:

Suppose $$S$$ is an ordered set and $$E\subset S$$. We say $$E$$ is bounded above iff: $$\exists \beta \in S, \beta \mbox{ is an upper bound of E}$$

Rudin chose to condense these two definitions into one. Does this clarify your question, or are you still confused?

• This explanation makes things a lot clearer for me, thank you! Is the empty set bounded above? Also, in this situation where S is an ordered set, B is empty, and L is the set of all lower bounds of B, would L be bounded above? Commented Jun 17 at 19:13
• The empty set is vacuously bounded. It’s not interesting. It’s not worth thinking about. $B$ is non empty in the theorem, so yes, $L$ is bounded above by any element of $B$. Please accept the answer, if you wish to do so. Commented Jun 17 at 19:35
• I accepted your answer! I know this is a weird case but is my thinking correct: if S=$\emptyset$ and B = $\emptyset$ we have $\emptyset \subset \emptyset$. We know that $\forall$ x $\in$ S $\forall$ y $\in$ B, y $\leq$ x. So all the x's in S are upper bounds of B. But there does not exist an x in S so B is not bounded above? Commented Jun 17 at 19:47
• If both $S$ and $B$ are empty, then $B$ is not bounded above (or below), and in particular does not have a least upper bound (or greatest lower bound). If, on the other hand, $B$ is empty but $S$ isn't, then $B$ is bounded above and below by every element of $S$. $B$ has a least upper bound if and only if $S$ has a smallest element, and $B$ has a greatest lower bound if and only if $S$ has a largest element. (And no, I didn't mistakenly interchange smallest amd largest in that last sentence.) Commented Jun 17 at 20:01
• Yes, if $B$ is empty, then all elements of $S$ are upper bounds for $B$. If $S$ has an element, then $B$ is bounded above, but not if $S$ is empty. Commented Jun 17 at 20:36

It's true that if we have any empty sets running around, we might have problems, but that's why $$B$$ is hypothesized to be nonempty and $$L$$ is shown to be nonempty. We know (by assumption of the theorem) that $$B$$ has elements in it, so saying that something is true for every $$x\in B$$ does mean that at least one $$x$$ meets the condition. (And in fact you can probably construct an example where if you let $$B$$ be empty, the conclusion doesn't hold.)

• Sounds good! Thank you! Commented Jun 17 at 20:53