$\sup{A} = \inf{B} \iff \forall \epsilon > 0, \, \exists a \in A, \, b \in B \, : \, b - a < \epsilon$

Question

If $A$ and $B$ are bounded subsets of $\mathbb{R}$, I need to show that, $\sup{A} = \inf{B} \iff \forall \epsilon > 0, \, \exists a \in A, \, b \in B \, : \, b - a < \epsilon$.

I have begun the proof in the "$\implies$" direction for now, but I am finding there are many cases. I ended up giving up a few times before I decided to ask here for help. Here is my proof so far.

$A$, $B$ bounded $\implies$ $\inf{A}$ and $\sup{B}$ exist. Also let $c = \sup{A} = \inf{B}$.

Without loss of generality, suppose $c - \inf{A} \leq \sup{B} - c$. That is, the "length" of $A$ is smaller than or equal to the "length" of $B$.

We will consider three cases for $\epsilon > 0$.

1. $\epsilon / 3 < c - \inf{A}$
2. $c - \inf{A} \leq \epsilon / 3 < \sup{B} - c$
3. $\epsilon / 3 \geq \sup{B} - c$.

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For case 1., let $a = c - \epsilon / 3 \in A$. Then let $b = c + \epsilon / 3 \in B$.

Then, $b - a = (2/3) \epsilon < \epsilon$.

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For case 2., let $a \in A$ be fixed, and let $b = a + \epsilon / 3 \in B$.

Then, $b - a = \epsilon / 3 < \epsilon$.

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I am unsure how to go about case 3.

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As you can see, case 3 is the difficult one to think about. If anybody could help me thinking of an $a \in A$ and $b \in B$ for case 3 that satisfied $b - a < \epsilon$ it would be greatly appreciated. Or... If someone can think of a better/more elegant proof, feel free to share.

Thanks.

Answer 1

It's somewhat easier to follow the following approach:

• Let $c=\sup(A)=\inf(B)$. Let $\varepsilon>0$.

• Since $c=\sup(A)$, there is some $a\in A$ such that $a>c-\varepsilon/2$.

• Since $c=\inf(B)$, there is some $b\in B$ such that $b<c+\varepsilon/2$.

Then, $b-a<\varepsilon$.

The other direction does not appear to be correct without additional assumptions, if $A=B=[0,1]$, the given conditions are trivially true since one can choose $a=b$, but the infimum and supremum are not equal. One way to correct the statement is to try to prove that:

$ \sup(A)\geq\inf(B)$ iff $\forall\varepsilon>0$, $\exists a\in A,b\in B$ such that $b-a<\varepsilon$.

Answer 2

For one direction we have $\epsilon>0$ Then we have $c(=sup{A}) + \epsilon/2$ is not a lower bound for $B$ so there exists $b\in B$ such that $b<c + \epsilon /2$. On the other direction use a similar argument to.conclude there exists $a\in A$ such that $a>c-\epsilon/2$. Use these two inequalities to get ur result.