If $A$ is a non-empty bounded subset of $\mathbb{R}$, and $B$ is the set of all upper bounds of $A$, prove that
$$glb(B)= lub(A)$$
My reasoning and thinking:
The set of upper bounds may be infinite countable set, right?
Also all the upper bounds (elements of set $B$) are greater than or equal to $x$ for all $x$ in $A$, using definition of upper bound.
Then it is $glb(B)$, I am not getting what does it mean? As $y$ is an element of $B$, then is $glb$ of $y$ is $y$ itself?
Any help/hint please.
Best Answer
First, observe that for any $b \in B$, $lub(A) \le b$, because $b \in B$ is an upper bound of $A$ and $lub(A)$ is the least of them. So $lub(A)$ is a lower bound of B, hence: $$lub(A) \le glb(B) \text{.} $$
Because $lub(A)$ is an upper bound of $A$, $lub(A) \in B$ by definition of $B$. Thus, $$glb(B) \le lub(A)\text{.} $$ Hence $lub(A) = glb(B)$.