Show that sets of real numbers $A, B$ are adjacent iff $\sup A = \inf B$

Question

If $A,B \subset \mathbb{R}$ satisfy : $$\begin{cases}\forall\ a \in A,\ \forall\ b \in B,\ a \le b \cr \forall\ \epsilon > 0,\ \exists\ a \in A,\ \exists\ b \in B \text{ such that }\quad b-a \le \epsilon\end{cases}$$ then we say that $A$ and $B$ are adjacent.

Show that $A$ and $B$ are adjacent if and only if : $\sup(A) = \inf(B)$.

My thoughts :

note that :

$$\sup A =: \begin{cases}\forall\ a \in A,\ ,\ a \le \sup A \cr \forall\ \epsilon > 0,\ \exists\ a \in A,\ \text{ such that }\quad \epsilon-\sup A < a\le \epsilon\end{cases}$$

$$\inf B =: \begin{cases}\forall\ b \in B,\ ,\ \inf B \le b \cr \forall\ \epsilon > 0,\ \exists\ b \in B,\ \text{ such that }\quad \inf B \le b < \inf B+\epsilon\end{cases}$$

• To show the first implication :

Assume that $A$ and $B$ are adjacent and let's show that $\sup(A)$, and $\inf(B)$ exists such that $\sup(A) = \inf(B)$.

• Show first that $\sup(A)$, and $\inf(B)$ exists :

Let $b\in B$, we have $$\forall a\in A,\quad a \le b$$ then b is upper bound, $A \neq \emptyset, A \subseteq \mathbb{R}$ then $\sup(A)$ exist.

Let $a\in A$, we have $$\forall b\in B,\quad a \le b$$ then a is Lower bound, $B \neq \emptyset, B \subseteq \mathbb{R}$ then $\inf(B)$ exist.

• Show first that $\sup(A)=\inf(B)$:

we have :$$\forall\ \epsilon > 0,\ \exists\ a \in A,\ \exists\ b \in B \text{ such that }\quad b-a \le \epsilon$$

or $$\forall\ \epsilon > 0,\ \exists\ a,b \in A\times B,\ \text{ such that }$$ $$ \begin{cases}\epsilon-\sup A < a\le \epsilon \cr \inf B \le b < \inf B+\epsilon\end{cases}$$ $$\iff \begin{cases}-\sup A < a\le \sup A-\epsilon\cr \inf B \le b < \inf B+\epsilon\end{cases}$$ $$\iff \inf B-\sup A < b-a < \sup A+\inf B $$ $$\iff \inf B-\sup A < \epsilon \quad \forall \epsilon > 0 $$ i'm stuk here

or we can say :

since $\forall a,b \in A\times B \quad a\leq b $ then $\forall b\in B,\quad \sup A \leq b$ then

$$\sup A \leq \inf B **(1)** $$

we have :$$\forall\ \epsilon > 0,\ \exists\ a \in A,\ \exists\ b \in B \text{ such that }\quad b-a \le \epsilon$$ then :$$\forall\ \epsilon > 0,\ \exists\ a \in A,\ \exists\ b \in B \text{ such that }\quad b<a+\epsilon $$ then :$$\forall\ \epsilon > 0,\ \exists\ a \in A,\ \exists\ b \in B \text{ such that }\quad \inf B \le \sup A + \epsilon $$

Then $$\inf B \le \sup A **(2)**$$

From (1) and (2) we have $$ \inf B=\sup A $$

• To show the second implication :

Assume that $\sup(A)$, and $\inf(B)$ exists such that $\sup(A) = \inf(B)$ and let's show that $A$ and $B$ are adjacent

• To show : $$\forall\ a \in A,\ \forall\ b \in B,\ a \le b $$

from the defintion of $\sup A$ and $\inf B$ we have: $$ \forall\ a \in A,\ ,\ a \le \sup A \text{ and } \forall\ b \in B,\ ,\ \inf B \le b$$ or we know that $\sup(A) = \inf(B)$ then $$\forall\ a \in A,\ \forall\ b \in B,\ a \le \sup A =\inf B\le b $$

• To show : $$ \forall\ \epsilon > 0,\ \exists\ a \in A,\ \exists\ b \in B \text{ such that }\quad b-a \le \epsilon $$

from the defintion of $\sup A$ and $\inf B$ we have: $$ \ \forall\ \epsilon > 0,\ \exists\ a \in A,\ \text{ such that }\quad \epsilon-\sup A < a\le \epsilon \text{ and } \forall\ \epsilon > 0,\ \exists\ b \in B,\ \text{ such that }\quad \inf B \le b < \inf B+\epsilon$$

we can also say that :

$$\forall\ \epsilon > 0,\ \exists\ a,b \in A\times B,\ \text{ tell que } \begin{cases}\frac{ \epsilon }{2}-\sup A < a\le \frac{ \epsilon }{2}\cr \inf B \le b < \inf B+\frac{ \epsilon }{2}\end{cases}$$ any help would be appreciated !!

Answer 1

HINT: If $\sup A<\inf B$ take $\epsilon<\inf B-\sup A$. To the reverse consider definition of $\sup$ and $\inf$ with the value $\epsilon/2$.

Answer 2

Your argument might be correct, but it is so clumsy that it takes about two printed pages. You may use freely the definitions of $\sup$ and $\inf$ as well as their standard properties, e.g., if $z<\sup A$ then there is an $a\in A$ with $z<a\leq\sup A$.

Let two nonempty sets $A$, $B\subset{\mathbb R}$ be given.

(I) If $\sup A=\sigma =\inf B$ then $a\leq\sigma\leq b$ for all $a\in A$ and all $b\in B$. Furthermore, for each $\epsilon>0$, there is an $a'\in A$ and a $b'\in B$ with $\sigma-{\epsilon\over2}<a'\leq\sigma\leq b'<\sigma+{\epsilon\over2}$. It follows that $b'-a'<\epsilon$; whence $A$ and $B$ are adjacent.

(II) If $A$ and $B$ are adjacent then $a\leq b$ for all $a\in A$, $b\in B$. It follows that any $b\in B$ is an upper bound for $A$, whence $\alpha:=\sup A\leq b$ for all $b\in B$. This says that $\alpha$ is a lower bound for $B$ and allows to conclude that $\alpha\leq\beta:=\inf B$. Now let an $\epsilon>0$ be given. There is an $a'\in A$ and a $b'\in B$ with $b'-a'<\epsilon$. It follows that $a'\leq \alpha\leq\beta\leq b'<a'+\epsilon$, and this implies $\beta-\alpha<\epsilon$. Since $\epsilon>0$ was arbitrary we can conclude that in fact $\alpha=\beta$.

Answer 3

Your proof starts off correctly by showing existence of $\sup A$ and $\inf B$ starting with assumption that $A, B$ are adjacent. To be frank, after this step you have lost somewhere in $\epsilon$'s. And it is better to point out the reason why you have lost like that.

Real-analysis is not shunting strange symbols like $\forall, \exists, \epsilon, \delta$ rather it is understanding order relations $(< , >)$ in the field of real numbers. Symbols like $\forall, \exists$ belong to "mathematical logic" and are more suitable for questions in that topic. Remember that ultimately all the proofs in mathematics are deep down an exercise in mathematical logic but looking in such manner at all proofs is not going to help at all.

Also another point about the textbook from which the question is taken. The definition of adjacency in question itself is full of these logical symbols and it clearly points out that the textbook is focusing more on symbol shunting than understanding of concepts. BTW its a typical issue with modern analysis texts (the justification being trying to fit more material in less number of pages).

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I rephrase the question as follows:

Two non-empty sets $A, B$ of real numbers are said to be adjacent if every member of $A$ is less than or equal to (or we can say "does not exceed") any member of $B$ and we can find a member of $A$ and a member of $B$ which are as close to each other as we want. Show that sets $A, B$ are adjacent if and only if $\sup A = \inf B$.

Suppose that $A, B$ are adjacent. Then like you have started with your solution we can prove that $\sup A, \inf B$ exist. We need to show that they are equal. Since every member of $A$ is less than or equal to any member of $B$ it follows that all members of $B$ serve as upper bound of $A$. Hence $\sup A$ is less than or equal to every member of $B$. It then follows that $\sup A$ is a lower bound for $B$ and hence $\sup A \leq \inf B$.

In order to show that $\sup A = \inf B$ we can use the use the technique of "proof by contradiction". Suppose on the contrary that $\sup A < \inf B$ and let $k = \inf B - \sup A$ so that $k > 0$. Then for every $a \in A$ and every $b \in B$ we have $a \leq \sup A < \inf B \leq b$. Thus $(b - a) \geq \inf B - \sup A = k > 0$. Hence we can not find a member of $A$ and a member of $B$ which are as close to each other as we want (their distance will always be greater than or equal to the positive number $k$). It follows that $A, B$ are not-adjacent. This contradiction shows that $\sup A = \inf B$.

To prove the other implication, suppose that $\sup A = \inf B$. Then for every $a \in A$ and every $b \in B$ we have $a \leq \sup A = \inf B \leq b$ so that the first condition for adjacency of $A, B$ is satisfied. Next note that by definition of $\sup A$ we can find a number $a \in A$ such that $a$ is as close to $\sup A$ as we want. Similarly we can find a number $b \in B$ which is as close to $\inf B$ as we want. Since $\sup A$ and $\inf B$ are same (say equal to $K$), we can choose numbers $a \in A, b \in B$ which are as close to a specific number $K$ as we want. Thus both the numbers $a, b$ must also be as close to each other as we want (Obviously if two numbers are close to a third number then they themselves can not be far apart. For if $|a - K|$ and $|b - K|$ are small then their sum is also small and $$|a - b| = |(a - K) + (K - b)| \leq |a - K| + |K - b|$$ so that $|a - b|$ is also small.) Thus the second condition for adjacency of $A, B$ is also established and therefore $A, B$ are adjacent.

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Note: The above formulation of the question and the corresponding answer avoids the symbol $\epsilon$ altogether but this does not lead to any loss of rigor. The symbol $\epsilon$ just represents a positive real number (and one can chose any other symbol like $a, b, c, d$ in its place) and its most common use in real-analysis is to quantify the distance between two real numbers. Thus the phrase "two numbers can be chosen as close to each other as we want" means that "for any $\epsilon > 0$ we can choose $a, b$ with $|a - b| < \epsilon$". If a student knows this proper meaning of the phrase "as close to as we want" then he may (and in my opinion perhaps should) avoid the use of $\epsilon$'s in solving small problems like these. $\epsilon$'s are not the substitute for "conceptual understanding".