Partially ordered set with no maximal element

Question

If I have a partially-ordered set $S$ (of some infinite cardinality), where for every $x$ in $S$ there is a $y$ in $S$ such that $x<y$. Is there a totally-ordered subset $C$ of $S$ satisfying these conditions?

• For every $x$ in $C$ there is a $y$ in $C$ such that $x<y$
• There is no $x$ in $S$ which is greater than every element of $C$

It seems to me intuitively that this is the case, but I can't figure out how to prove it.

Answer 1

Since $S$ is of infinite cardinality, it is non-empty, so $\exists~x_0\in S$

Now, consider $x_1\in S$ such that $x_0\lt x_1$. Continuing with $x_k\in S$, we consider $x_{k+1}\in S$ such that $x_k\lt x_{k+1}$ ad infinitum.

Now, define $C:=\{x_i\}_{i\in I}$ where $I$ is an index set with cardinality $|S|$ (a proper subset $A\subset S$ exists such that $|A|=|S|$ since $S$ is infinite, we take $I$ to be the index set of $A$).

This is totally ordered since for any $x_i,x_j\in C$, we have $x_i\lt x_j$ for $i\leq j$

• The first condition is obviously satisfied, since for $x_n\in C$, we have $x_{n+1}\in C$ such that $x_n\lt x_{n+1}$

• For the second one, suppose there exists $y\in S$ such that $x\lt y~\forall~x\in C$. But, by the construction of $C$, it has no maximal element, a contradiction to the existence of $y$.

Answer 2

Partial answer. Assume that $S$ is a countable partially-ordered set with no maximal element. Say $S=\{s_0,s_1,s_2,\dots,s_n,\dots\}.$ Now define a sequence $c_0,c_1,c_2,\dots,c_n,\dots$ recursively, as follows. Start with any element $c_0\in S.$ Having defined $c_n,$ let $c_{n+1}=s_n$ if $s_n\gt c_n;$ otherwise, choose any $c_{n+1}\in S$ so that $c_{n+1}\gt c_n.$ Finally let $C=\{c_0,c_1,c_2,\dots,c_n,\dots\}.$ Since $c_0\lt c_1\lt c_2\lt\cdots,$ it follows that $C$ is totally ordered and satisfies your first condition. To see that the second condition is satisfied, consider any element $s\in S,$ and assume for a contradiction that $s$ is greater than every element of $C.$ Then $s=s_n$ for some $n.$ Since $s_n\gt c_n,$ it follows that $s_n=c_{n+1}\lt c_{n+2}\in C.$

That answers your question for countable sets. For the general case, you will need some form of the axiom of choice, such as Zorn's Lemma or the Well Ordering Theorem.

Answer 3

Any maximal chain C, will suffice for if C had a maximum,
then S would have a maximal element, which it does not.