# Existence of coinitial and cofinal sequences on uncontable and totally ordered set

Let be $\Omega$ an uncountable set and $\leq$ a linearly ordered relation on $\Omega$. Than can we say that exists a co-initial sequence and cofinal sequence in $\Omega$?

Where we say:

a subset $B$ of $\Omega$ is said to be coinitial if it satisfies the following condition: For every $a \in \Omega$, there exists some $b \in B$ such that $b \le a$.

A subset $B$ of $\Omega$ is said to be cofinal if it satisfies the following condition: For every $a ∈ \Omega$, there exists some $b ∈ B$ such that $a \leq b$.

• Yes.... $\Omega$ is such a set. – Stefan Mesken Mar 12 '16 at 10:59
• Thanks @Stefan, can you also provide some evidence of why $\Omega$ is such a set? – Sara Mar 12 '16 at 11:10
• We have $a \le a$ for every $a \in \Omega$. We may thus let $b := a$ for every $a \in \Omega$ to witness that $\Omega$ is both coinitial and cofinal. – Stefan Mesken Mar 12 '16 at 11:50
• If by sequence you mean indexed by an ordinal, then yes. If by sequence you mean indexed by $\Bbb N$, then no. – Asaf Karagila Mar 12 '16 at 12:09
• @AsafKaragila, I'm considering a sequenced indexed by an ordinal. Suppose we have already proof that all bounded intervals in $\Omega$ have cardinality at most $m$. $\aleph_{0}\le m\le |\Omega|$. Now let $(\alpha_{\xi})_{\xi < \nu_{0}}$ be a co-initial sequence in $\Omega$ and let $(\beta_{\mu})_{\mu < \nu_{1}}$ be a cofinal sequence, where $\nu_{0},\ \nu_{1}$ are suitable ordinals and every proper initial segment of $\nu_{0}$ has cardinality at most $m$ etc. I don't understand why I can consider these two sequences and why the proper initial segment of $\nu_{0}$ has cardinality at most $m$. – Sara Mar 12 '16 at 13:41

## 1 Answer

This is easiest using transfinite recursion.

Fix a well-ordering of $\Omega$, now proceed by picking a random element, then at each step if you're still bounded, pick a larger/smaller element. Until you've exhausted all the elements or got an unbounded sequence.

You can also use Zorn's lemma when you order well-ordered (or anti-well ordered) subsets by end-extension ($A\unlhd B$ if $B\cap(-\infty,\sup A)=A$). It is not hard to show the conditions for Zorn's lemma hold, and therefore there is an unbounded subset of $\Omega$ which is well-ordered under $\leq$.

The axiom of choice is necessary, by the way, as it is consistent that there is a subset of $\Bbb R$ which is unbounded, but has no countably infinite subset. Looking at the induced linear order from $\Bbb R$ you get an uncountable set, with a linear order, but every sequence is finite and therefore bounded.