Let $A$ be an infinite set that includes Real numbers and is bounded. Let $B$ be a set of Real numbers $x$ s.t. the intersection $A\cap[x,\infty)$ is empty or includes $finite$ number of elements.
prove that $\inf B$ exists.
prove or disprove $\inf B=\min B$
prove or disprove that $\inf B$ exists if we don’t ask for $A$ to be bounded.
I started searching for examples to see the whole picture, like sequences of the form “$\frac{1}{n}”$ and others, but I didn’t know to to do the intersection and why I must be sure that B has an infimum.
its easier if someone can give an example of two sets $A$ and $B$, and make it clear why the intersection must be a finite set. (after choosing “x” from set B)
Hint: if you do not impose finiteness of the intersection, then you essentially lose control on $B$, which can be anything you want since the intersection condition is vacuous now.
if $\inf B=-\infty$, then $[x,+\infty)\cap A = A$ for some $x$ since $A$ is bounded…
$A=(0,1)$, $B=(5,6)$…
$A:=\{-1,-2,-3,-4,\cdots\}$, $B=\mathbb{R}$