Let I be a totally ordered set(偏序集), $E\subsetneq I$ be a subset of I, Infimum of E is the greatest lower bound of $E$

the least upper bound

$$ \inf E,\sup E $$

if $\inf E\in E$, $\min E=\inf E$

proposition:

1. $\inf(E\cup F )=\min (\inf E,\inf F)$
2. $E\subset F$, ⇒ $\inf E\ge \inf F$
3. intersection $\inf(E\cap F)\ge \max (\inf E,\inf F)$
4. minkfusik $\inf(E+F)=\inf E+\inf F$
5. scalar $\inf tE=t\inf E$ when $t\ge 0$
6. $\inf -E=\sup E$

real-value function$f:X\to \R$ ←ordered subset $X$

$$ \inf_{x\in X}f(x)=\inf \{f(x):x\in X\}\in\R $$

inf of range of function of f

proposition: Let $f_i=\inf_{x\in X}f(x)$, then

1. $\inf_{x\in X_1\cup X_2}=\min(f_1,f_2)$

2. if $X_1\subset X_2$ , then $f_1\ge f_2$

3. $\inf_{x\in X_1\cap X_2}\ge \max (f_1,f_2)$

4. $\inf \{f(x_1)+f(x_2),x_1\in X_1,x_2\in X_2\}=f_1+f_2$. just plus useful→

   $$ \inf f(x_1)+g(x_2)=\inf_{x\in X_1} f(x)+\inf_{x\in X_2} g(x) $$

   x in right is independent between f & g, which is not below:

   $$ \inf (f+g)\ge \inf g+\inf f $$

5. $\inf _{x\in X}tf=tf,t\ge 0$