Prove of disprove this statement: Let A be a subset of $\mathbb{R}$ which is nonempty and bounded above, and let
$B=\{b\in \mathbb{R} : b\space is\space an\space upper\space bound\space for\space A\}$
Then inf B= sup A
Intutively, I think this statement is true. Here is my argument.
By Completeness Axiom, supA exists, which means the least upper bound of A exists.
Then we know that the minimum of B exists.
Moreover, we have $infB=minB$=$supA$
I wonder whether this argument is correct, or if this statement is wrong, can anyone provide a counterexample?
Thanks in advance!
You're not wrong.
By definition, the $\sup_{a \in A}$ is the smallest upper bound of $A$. Which means it's the smallest member of $B$ or $\min_{b \in B}$. Which is the biggest lower bound of $B$ or $\inf_{b \in B}$. So we have. $$\sup_{a \in A} = \min_{b \in B} =\inf_{b \in B}$$
A simple example would be for $A = (-\infty,0)$ and $B=[0,\infty)$
clearly, if $b \in B$ then $b$ is an upper bound of $A$. And $\sup_{a \in A} = \inf_{b \in B} = 0$