Question on Partially ordered sets and images of sets

Question

Could someone look through my attempt at proving the following problem please?

Let $(A,\preceq)$ and $(B,\preceq')$ be POSETS and $C \subseteq A$. Suppose that $h:A \rightarrow B$ satisifies $x \preceq y \iff h(x) \preceq' h(y)$ for all $x,y \in A$.

If $h[C]$ has an upper bound and $h$ is onto, then prove that $C$ has an upper bound.

---

Attempt :

Let $(A,\preceq)$ and $(B,\preceq')$ be POSETS and $C \subseteq A$. Suppose that a function $h:A \rightarrow B$ satisifies $x \preceq y \iff h(x) \preceq' h(y)$ for all $x,y \in A$.

Suppose for contradiction that $C$ does not have an upper bound then by definition, there does not exist any $b \in A$ such that $x \preceq b$ for any $x \in C$. [i]

By assumption, $h[C]$ has an upper bound, therefore by definition there exists some $d_B \in h[C]$ (with $d_B=h(d_A)$ for some $d_A \in C$) such that $z \preceq' d_B$ for all $z \in h[C]$ (with $z=h(x)$ and $x \in C$).

Since $z \preceq' d_B$ and $d_B=h(d_A)$ and $z=h(x)$ then $h(x) \preceq' h(d_A)$ and $x \preceq d_A$ by definition, which contradicts the result found in [i].

---

Again...this looks right to me...but feels iffy for some reason...i did not even use the fact that h is onto!

Thanks again.

Answer 1

Your proof is right, but you do use that $h$ is onto. Namely when you say $d_B = h(d_A)$ for some $d_A \in A$.

Also, you do not need to work with a contradiction. The middle part of your proof directly proves that there is some upper bound for $C$ in $A$. Right now, your proof is of the form:

1. Assume $C$ has no upper bound in $A$.
2. Show that $C$ has an upper bound in $A$, without using the assumption in the first step.
3. This contradicts the first assumption, so contradiction, so $C$ must have an upper bound (which we already directly established in the second step).

Fun extra exercise: can you think of an example of $A, B, C$ and $h$ that satisfy all the hypothesis except that $h$ may not be onto, such that $h[C]$ has an upper bound in $B$ but $C$ has no upper bound in $A$.