$\sup A < \inf B$ implies existence of $c \in \mathbb{R}$ such that $a < c < b$

Question

Related: What are necessary and sufficient conditions such that sup A < inf B implies there is a c such that a<c<b for all a,b in their respecitve sets

Working through some revision Analysis questions and got stumped on this. Asked to determine whether the following statement is true or not (and if not, a counterexample);

If $\sup A < \inf B$ for sets $A, B$ in $\mathbb{R}$, then there exists $c \in \mathbb{R}$ such that $ a < c < b$ for all $a \in A$, and $b \in B$.

An answer to the above question linked indicates this is true, but I haven't gotten any further to proving it other than noting that $\forall \epsilon > 0$, there are $a$ and $b$ such that $\sup A < a + \epsilon$ and $b-\epsilon < \inf B$. Hints?

You need to have $\sup A<\inf B$ otherwise this is not possible. — Paramanand Singh, Apr 28 '17 at 05:31
Ok note that if $c$ is any number such that $\sup A<c<\inf B$ then it has the desired property. — Paramanand Singh, Apr 28 '17 at 07:02
Sure that's fine, but how do I construct $c$? Can I use $a+\epsilon$ and $b-\epsilon$ and some kind of sequence argument? — Adrian Hindes, Apr 28 '17 at 07:09
why construct $c$? Any number $c$ between $\sup A$ and $\inf B$ works and there are infinitely many such numbers. Remember, between any two distinct real numbers there lie an infinity of real numbers. Most of the analysis questions don't need anything related to $\epsilon$ gymnastics. — Paramanand Singh, Apr 28 '17 at 07:11

score 3 · Answer 1 · answered Apr 28 '17 at 05:25

3

This is clearly wrong. Take $A=[0,1]$ and $B=[0,2]$. As $1 \in A > 0 \in B$, there can not exist such a $c$.

answered Apr 28 '17 at 05:25

Vincent

618

Ah thanks, didn't catch the typo, it's meant to be $\sup A < \inf B$ – Adrian Hindes Apr 28 '17 at 06:44

Paramanand Singh · Accepted Answer · 2017-04-28T07:28:25.387

After the typo correction by OP, the question is trivial.

If $A, B$ are not empty sets of real numbers such that $\sup A < \inf B$ then there exists a real number $c$ with the property that $a < c < b$ for all $a\in A, b\in B$.

Why is this trivial? Simply because we can take any number $c$ such that $\sup A < c <\inf B$ (such a $c$ exists because given any two distinct real numbers there lie an infinity of real numbers between them). And then for all $a \in A, b\in B$ we have $$a\leq \sup A < c < \inf B \leq b$$ by definition of $\inf, \sup$.

Note that if $\sup A \nless \inf B$ then such a $c$ can't be found and it is easy to construct example sets $A, B$ showing the non-existence of such a $c$.

Also understand that most results and theorems in (introductory) analysis use the density of real numbers and the results of these kind remain valid in rational number system also and such results can be considered trivial (i.e based on direct application of definitions). Most of the problems involving $\epsilon-\delta$ gymnastics also fall under this category and are made difficult only by presence of greek symbols. The really significant results of analysis deal with the completeness property of real numbers and they can not be solved using common sense or algebraic techniques. For such problems you need to have an understanding of order relations and completeness of real numbers.

Ah thanks for that! Don't know why it didn't click that of course there is some numbers between! — Adrian Hindes, Apr 28 '17 at 08:34

score 0 · Answer 3 · answered Apr 28 '17 at 05:53

0

it is not true: A={1, 5, 10} B={1, 3, 15}. supA=10 supB=15. So far all conditions are met.

Now choose a=1 ;b=1. And the staetement is not possible.

answered Apr 28 '17 at 05:53

dvd280

676

$\sup A < \inf B$ implies existence of $c \in \mathbb{R}$ such that $a < c < b$

3 Answers3