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.
Best Answer
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 $y$.