I am trying to show that given $A, B \subset R$, both non-empty, show that $\inf{(A \cup B)} = \min{\{\inf{A}, \inf{B}\}}$. I have done it using definitions mainly but I was wondering if there was a more rigorous approach to take:
By definition, $\inf A \le a$ $\forall a \in A$ and $\inf B \le b$ $\forall b \in B$, from this it is clear that if we combine $B$ and $A$, its infimum must be less than $\min{\{\inf A, \inf B \}}$ for $\inf (A \cup B) \le x$ $\forall x \in A \cup B$ to hold by definition.
Any advice or confirmation of this proof being sufficient is appreciated, thanks!
Best Answer
It is always true that if $C \supset D$ then $\inf C \le \inf D$ so you have $\inf A\cup B \le \inf A$ and $\inf A\cup B \le \inf B$ which gives $\inf A \cup B \le \min ( \inf A, \inf B )$.
If $x \in A \cup B$, then either $x \in A$ or $x\in B$. Then $x \ge \inf A$ or $x \ge \inf B$ and so $x \ge \min ( \inf A, \inf B )$, from which it follows that $\inf A \cup B \ge \min ( \inf A, \inf B )$.