# Thread: Immediate successors in Well-ordered sets

1. ## Immediate successors in Well-ordered sets

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 $\leq$. Let $x \in X$ be arbitrary but fixed.

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

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

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

Questions:

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

2. How do I show (*)?

3. How do I show (**)?

2. 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 $\leq$. Let $x \in X$ be arbitrary but fixed.

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

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

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

This isn't usually true

i.e. there doesn't exist a $y \in X$ such that $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 $S:= \{s\in X\;;\;x\leq s\}$ , 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

3. Originally Posted by tonio
Look at the set $S:= \{s\in X\;;\;x\leq s\}$ , 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!