let $(X, \leq_*)$ be a partially ordered set.

Assume there is an isomorphism $f: (X,\leq_*) \to (\mathbb Z, \leq)$

let $A \subseteq X$ be a well ordered subset of $X$ with an upper bound. Meaning there is an element $b \in X$ such that $\forall a \in A, a\leq_* b$.

Show that $A$ is a finite set.

I'm not sure how to approach this problem. And I'm not sure I understand what we are given too. on one hand, $(X,\leq_*)$ is partially ordered. But on the other hand, it is isomorphic to $(\mathbb Z,\leq)$ which is linearly ordered.


If $f$ is an isomorphism, you might as well assume that $X=\Bbb Z$. It's easy to prove this there, if $A$ is a well-ordered subset of $\Bbb Z$ and it has an upper bound, then it must be a subset of some interval $[-k,k]$ and so finite.

Returning to the general case, if $f$ is an isomorphism, then well-ordered sets go to well-ordered sets, and bounded sets go to bounded sets, therefore the image of $A$ under $f$ is finite, so $A$ is finite.

  • $\begingroup$ Yes I see now. the image of $A$ under $f$ is a subset of $\mathbb Z$, and it has an upper and lower bound so it is finite, and since $f$ is an ismorphism, $A$ is finite. Just one question, isn't it a contradiction to say that $(X,\leq_*)$ is partially ordered but it is isomorphic to a linearly ordered set? $(\mathbb Z,\leq)$ $\endgroup$ – Oria Gruber Feb 1 '14 at 14:06
  • $\begingroup$ Linear orders $\subsetneq$ Partial orders. $\endgroup$ – Asaf Karagila Feb 1 '14 at 14:09

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