Given two bounded sets $A,B$ and $\sup A<\sup B$, is there an element in $B$ that works as an upper bound for $A$?

Question

Originally I worked this questions out by just saying that $m=\sup B$ and $m>\sup A$. Then drawing the conclusion that $m$ is an upper bound for $A$. I figured this was wrong because we're unsure that $\sup B$ is in $B$ at all.

I decided to approach it like this:

Let $\sup B=m$, which may or may not be in $B$. But, $m-1$ is in $B$ and $m-1\geq \sup A$. Therefore, $m-1$ is an upper bound for $A$.

My main question arises when subtracting $1$ from $m$. Is this a logical justification? I can't think of any other way to justify it unless I know that the set $B$ has maximum, which I don't.

Thank you.

Answer 1

You can't necessarily subtract $1$ and stay larger that $\sup A$. But you can subtract anything that is smaller than $\varepsilon=|\sup A - \sup B|$.

By definition of a supremum, $B$ contains an element in the interval $(m-{\varepsilon\over2}, m)$ and that element is an upper bound for $A$.

Answer 2

Pick an $\epsilon>0$ such that $\sup A<\sup B-\epsilon$, then there exists some $b\in B$ such that $\sup B-\epsilon<b\leq\sup B$, then $b>\sup A\geq a$ for all $a\in A$.

Answer 3

No, you can't subtract $1$ like that. But $\sup B$ is the least upper bound for $B$; since $\sup A<\sup B$ this says that $\sup A$ is not an upper bound for $B$, which says in turn that there exists $b\in B$ with $b>\sup A$ (and hence yes, $b$ is an upper bound for $A$).

Answer 4

$\sup B$ is the smallest upper bound for $B$. This means that a strictly smaller number $\sup A$ cannot be another upper bound of $B$. Thus, $\mathbb R$ being totally ordered, there is a $b\in B$ such that $b\gt\sup A$. However, this means that, for every $a\in A$ you have $b\gt\sup A\gt a$, i.e. $b$ is an upper bound for $A$.

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The condition about total ordering cannot be dropped. Imagine the set $\mathbb Z\cup\{a,b\}$ such that the relation $\le$ is given by:

$$x\le y\Longleftrightarrow\begin{cases}x\le y\text{ as integers}&x,y\in\mathbb Z\\\text{true}&y=b\\\text{true}&x=y\\\text{false}&\text{otherwise}\end{cases}$$

In other words, $b$ is the maximum of the whole set, on one hand bigger than all integers, on the other bigger than $a$, but the integers are not comparable with $a$. Now, take $A=\{a\}, B=\mathbb Z$ and you will see that $a=\sup A, b=\sup B$ and $b\gt a$ but none of the elements of $\mathbb Z$ is comparable to $a$ so cannot be an upper bound of $A$.

Answer 5

No, it's not justified, unless $A$ and $B$ are sets of integers. What you can say,is that $m -\frac12(m-\sup A)>\sup A$, and there's necessarily an element $b\in B$ such that $$m\ge b>m-\tfrac12(m-\sup A)>\sup A.$$ since $m=\sup B$. The conclusion follows.

Answer 6

This is an answer to the question in the title of question and preassumes that we are dealing with a total order that has the least upper bound property.

If no such $b\in B$ exists then automatically $\sup A$ is an upper bound of $B$ so that $\sup B\leq\sup A$. This contradicts the data in your question.

Answer 7

By definition any number greater than $\sup A$ is an upper bound for $A$ and thus every number lying in interval $[\sup A, \sup B] $ is an upper bound for $A$. And by definition of supremum there exists at least one element $b\in B$ which lies in interval $(\sup A, \sup B] $. Thus $b\in B$ is an upper bound for $A$.