A little help with this question?

Let be a well-ordered set. Show that for each , either is the greatest element of or has an immediate successor (that is, there exists such that there does not exist with )

Now, here's how I started:

Let X be well-ordered by $\displaystyle \leq $. Let $\displaystyle x \in X $ be arbitrary but fixed.

If x is the greatest element of X, then $\displaystyle \forall y \in X, y \leq x $.

If x is not the greatest element of X, then $\displaystyle x^+ \in X $ (*) and $\displaystyle x \leq x^+ $.

Then I claim that $\displaystyle x^+ $ is the immediate successor of x, i.e. there doesn't exist a $\displaystyle y \in X $ such that $\displaystyle x<y<x^+ $ (**)

Questions:

1. Does the proof depend on the choice of well-ordering?

2. How do I show (*)?

3. How do I show (**)?

Originally Posted by Mimi89

A little help with this question?

Let be a well-ordered set. Show that for each , either is the greatest element of or has an immediate successor (that is, there exists such that there does not exist with )

Now, here's how I started:

Let X be well-ordered by $\displaystyle \leq $. Let $\displaystyle x \in X $ be arbitrary but fixed.

If x is the greatest element of X, then $\displaystyle \forall y \in X, y \leq x $.

If x is not the greatest element of X, then $\displaystyle x^+ \in X $ (*) and $\displaystyle x \leq x^+ $.

Then I claim that $\displaystyle x^+ $ is the immediate successor of x,


This isn't usually true


i.e. there doesn't exist a $\displaystyle y \in X $ such that $\displaystyle x<y<x^+ $ (**)

Questions:

1. Does the proof depend on the choice of well-ordering?

2. How do I show (*)?

3. How do I show (**)?

Look at the set $\displaystyle S:= \{s\in X\;;\;x\leq s\}$ , and assuming that x isn't the greatest element of X and that X i well-ordered take now a first element in S ...

Tonio

Originally Posted by tonio

Look at the set $\displaystyle S:= \{s\in X\;;\;x\leq s\}$ , and assuming that x isn't the greatest element of X and that X i well-ordered take now a first element in S ...

Tonio

Thank you!!! Though it might seem obvious to you, it didn't occur to me to look at this set and say it must have a minimal element!