# Segments, posets

• Mar 7th 2010, 12:42 PM
vaevictis59
Segments, posets
Prove that for all $\alpha$ in $\omega_1$,

$\text{seg}_{\omega_1}(\alpha) +_o \omega_1 =_o \omega_1$

and

$
\text{seg}_{\omega_1}(\alpha) \cdot_o \omega_1 =_o \omega_1$
.

Notation: $\omega_1$ denotes the first uncountable ordinal. $\text{seg}$ above denotes the initial segment, $\cdot_o$ denotes the product of posets, $+_o$ denotes the sum of posets. $U <_o V \Leftrightarrow (\exists x \in V ) [U =_o \text{seg}_V(x)]$. $=_o$ denotes order isomorphic.