Finding an ordered set

Question

Find a partially ordered set $A$ so that for every member a in $A$, there is $b∈A$ such that $b<a$ but for infinitely many members of $A$, there does not exist $c\in A$ such that $a<c$.

Thanks.

Answer 1

Let $A$ be the set of positive integers, and write $x\lt y$ if $y$ divides $x$ but $y\ne x$. Informally, the condition says that $x$ is a proper multiple of $y$. This is the "natural" divisibility relation on the natural numbers, just written backwards.

Certainly for any $a$ there is a $b$ such that $b\lt a$, that is, $b$ is a proper multiple of $a$.

But if $a$ is prime or $1$, there is no $c$ such that $a\lt c$. And there are infinitely many primes.

Remark: If one is less number-theoretically minded, let $A$ be the collection of all finite subsets of an infinite set $I$. If $x$ and $y$ are such subsets, write $x\lt y$ if $y$ is a proper subset of $x$.