Maximal exercise (Set Theory).

Question

Let $A$ and $B$ be partially ordered sets, and let $f:A→B$ be strictly increasing function. Prove that if $b$ is maximal element of $B$ , then each of $f^*(b)$ is a maximal element of A.

I tried to reason by absurdity, I assumed that $ f ^ * (b) $ is not a maximal element of $ A $, but I cannot compare an element of $ a $ with a subset of $ A $, since $ f ^ * ( b) = f ^ {- 1} (\{b\}) $ ($ f ^ *(b)$ is a subset of $ A $) I don't know how to compare $ f ^ * (b) $ with some element of $ A $ to use that $ f $ is increasing.

Answer 1

"but I cannot compare an element of a with a subset of A"

Yes, you can.

Because one element of $a\in f^*(b)$ can be a representative element. Prove that if $a\in f^*(b)$ then $a$ must be maximal in $A$.... As $a$ was utterly arbitrary in $f^{*}(b)$ it must be true that all elemetns of $f^*(b)$ are maximal.

And, yes, a proof by contradiction is an excellent way to go.

Assume $a\in f^*(b)$ so $f(a) = b$. And assume $a$ is not maximal in A. So there is a $c \in A$ so that $c > a$.

Then......

$f(c) > f(a)=b$ because $f$ is increasing. So $f(c) > b$.

Answer 2

Take $x \in f^*(b)$. If $x \leq y$, our goal is to prove that $x=y$. Apply $f$ to get $b \leq f(y)$. Since $b$ is maximal, $b =f(y)$. Now, $x\leq y$ means that either $x<y$ or $x=y$. The first case is impossible, since $f$ being strictly increasing would lead to $b<b$. Thus it must be $x=y$.