## Franklin's Notes

### Separable order type

Given an ordered set $A$ and a subset $B\subset A$, we say that $B$ is dense in $A$ iff for all elements $a_1<a_2$ of $A$, there exists $b\in B$ such that $a_1<b<a_2$. That is, between any two distinct elements of $A$, there exists an element of $B$, so that $B$ is "ubiquitous" throughout $A$ in some sense. If $B$ is both countable and dense, then we say that $A$ is separable. Similarly, we say that an order type $\alpha$ is separable if it is the order type of a separable ordering.

Proposition 1. If $A$ is separable and ${I_j}_{j\in J}$ is a condensation of $A$ in which no $I_j$ is a singleton, then $J$ is at most countable. That is, $A$ can be partitioned into at most countably many (nontrivial) intervals.

Proof. We shall explicitly define an injection $f:J\to\mathbb N$, given that $A$ is separable and ${I_j}_{j\in J}$ is a condensation with nontrivial components. If $B\subset A$ is a countable dense subset, then we have a bijection $b:\mathbb N\to B$. Then we may define $f(j)$ to be the least $n\in\mathbb N$ such that $b(n)\in I_j$. This is well-defined since each interval $I_j$ is not a singleton, meaning that each one contains two distinct elements, and some element of $B$ must lie between these two elements and therefore also in $I_j$, so that such a natural number $n$ exists for all intervals $I_j$. This is an injection because the intervals are a partition, meaning that no $b(n)$ can be in two of the intervals simultaneously. Hence, since we have an injection $f:J\to \mathbb N$, it follows that $|J|\leq |\mathbb N|$ as claimed. $\blacksquare$

We have the following corollary as an application of this proposition, which gives rise to a useful insight about the ordering on the real numbers:

Proposition 2. The real numbers cannot be partitioned into uncountably many intervals. Consequently, $\omega_1$, the first uncountable ordinal , cannot be embedded in $\lambda$, the order type of the real numbers.

Proof. Notice that $\mathbb Q\subset\mathbb R$ is a countable dense subset of $\mathbb R$, so that $\mathbb R$ is separable. Consequently, any partition of $\mathbb R$ into intervals must consist of at most countably many intervals. Any embedding of an infinite ordinal in $\mathbb R$ gives rise to a partition of $\mathbb R$ into a set of intervals with the same cardinality, from which it follows that no uncountable ordinal can be embedded in $\mathbb R$. $\blacksquare$