# Relation between open bounded intervals and max, min of two real numbers.

In a textbook on real analysis, they define max and min of any two real numbers $$a,b$$ as follows:-

$$\max\{a,b\}= \dfrac{1}{2}(a+b+|a-b|)$$

$$\min\{a,b\}= \dfrac{1}{2}(a+b-|a-b|)$$

Consider two open bounded intervals, $$(a_1, a_2)$$ and $$(a_2, b_2)$$. Now let $$c\in \mathbb{R}$$ and $$c\in (a_1, b_1)$$ and also $$c\in (a_2, b_2)$$.

Let $$a_3= \min\{a_1, a_2)$$ and $$b_3= \max\{b_1, b_2\}$$.

Now they say $$(a_3, b_3)= (a_1, b_1)\cup (a_2, b_2)$$ and also $$c\in (a_3, b_3)$$.

(i) What is the proof for $$(a_3, b_3)= (a_1, b_1)\cup (a_2, b_2)$$ given the above conditions.

• I think you meant $(a_1,b_1)$ and $(a_2,b_2)$, isn't it? If $(a_1,a_2)$ is non empty $a_1 < a_2$ so $a_3 = a_1$. Commented May 18, 2021 at 9:31

We know that $$(a,b) \cap (c,d) = (\max(a,c) , \min(b,d))$$ so $$((a_1,b_1) \cup (a_2,b_2))^c = (a_1,b_1)^c \cap (a_2,b_2)^c = [(-\infty,a_1) \cup (b_1,+\infty)]\cap [(-\infty,a_2,) \cup (b_2,+\infty)]$$

$$= (-\infty,\min(a_1,a_2)) \cup (\max(b_1,b_2),+\infty)$$ where I used the fact that $$\emptyset \ne (a_1,b_1) \cap (a_2,b_2) = (\max(a_1,a_2),\min(b_1,b_2))$$ yields $$\max(a_1,a_2) < \min(b_1,b_2)$$. Now $$(a_1,b_1) \cup (a_2,b_2) = [(-\infty,\min(a_1,a_2)) \cup (\max(b_1,b_2),+\infty)]^c = (a_3,b_3).$$

• One would then question how do you "know" $(a,b)\cap(c,d)=(\max\{a,c\},\min\{b,d\})$ Commented May 18, 2021 at 9:47
• That is a direct consequence of the definition of an interval. Commented May 18, 2021 at 15:11

First note that $$\exists x|x\in(a_1,b_1)\text{ and }x\in(a_2,b_2)\rightarrow b_2>a_1\text{ and }b_1>a_2$$ which is due to the fact that $$a_1,a_2.

$$(a_1,b_1)^c=\{x\in\Bbb R:x\le a_1\text{ or }x\ge b_1\}$$ and similarly $$(a_2,b_2)^c$$.

\begin{align*}&x\in(a_1,b_1)^c\cap(a_2,b_2)^c\\ &\leftrightarrow(x\le a_1\text{ or }x\ge b_1)\text{ and }(x\le a_2\text{ or }x\ge b_2)\\ &\leftrightarrow (x\le a_1\text{ and }x\le a_2)\text{or}\color{red}{(x\le a_1\text{ and }x\ge b_2)}\text{or}\color{red}{(x\ge b_1\text{ and }x\le a_2)}\text{or}(x\ge b_1\text{ and }x\ge b_2) \end{align*}

You can verify that $$(x\le a_1\text{ and }x\le a_2)\leftrightarrow x\le\min\{a_1,a_2\}\\ (x\ge b_1\text{ and }x\ge b_2)\leftrightarrow x\ge\max\{b_1,b_2\}$$and the two red cases are empty sets as $$b_2>a_1$$ and $$b_1>a_2$$. So

$$x\in(a_1,b_1)^c\cap(a_2,b_2)^c\leftrightarrow x\le\min\{a_1,a_2\}\text{ or } x\ge\max\{b_1,b_2\}$$and by De Morgan's Law$$x\in[(a_1,b_1)^c\cap(a_2,b_2)^c]^c=(a_1,b_1)\cup(a_2,b_2)\leftrightarrow x>\min\{a_1,a_2\}\text{ and }x<\max\{b_1,b_2\}$$

which is equivalent to $$x\in(a_3,b_3).~\blacksquare$$