# uncountable set

• Sep 15th 2009, 05:02 PM
dori1123
uncountable set
Let $\displaystyle S_\omega$ be the minimal uncountable well-ordered set.
(a) Show that $\displaystyle S_\omega$ has no largest element.
(b) Show that for every $\displaystyle \alpha \in S_\omega$, the subset $\displaystyle \{x | \alpha < x\}$ is uncountable.
(c) Let $\displaystyle X_0$ be the subset of $\displaystyle S_\omega$ consisting of all elements $\displaystyle x$ such that $\displaystyle x$ has no immediate predecessor. Show that $\displaystyle X_0$ is uncountable.

I know how to do (a) and (b) but don''t know (c), can anyone help?
• Sep 20th 2009, 09:45 AM
Failure
Quote:

Originally Posted by dori1123
Let $\displaystyle S_\omega$ be the minimal uncountable well-ordered set.
(a) Show that $\displaystyle S_\omega$ has no largest element.
(b) Show that for every $\displaystyle \alpha \in S_\omega$, the subset $\displaystyle \{x | \alpha < x\}$ is uncountable.
(c) Let $\displaystyle X_0$ be the subset of $\displaystyle S_\omega$ consisting of all elements $\displaystyle x$ such that $\displaystyle x$ has no immediate predecessor. Show that $\displaystyle X_0$ is uncountable.

I know how to do (a) and (b) but don''t know (c), can anyone help?

Claim: If $\displaystyle Y$ is subset of $\displaystyle S_\omega$ in which every element has an immediate predecessor it follows that $\displaystyle Y$ is countable.

Thus, given any $\displaystyle \alpha\in X_0$ the largest subset of $\displaystyle S_\omega$ that contains $\displaystyle \alpha$ and in which every element has an immediate predecessor is countable.
It follows that if $\displaystyle X_0$ were countable, then $\displaystyle S_\omega$ would be a countable union of countable sets, hence countable.