Immediate successors in Well-ordered sets

**Mimi89** · Mar 9th 2010, 02:22 AM

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 $x<y<x^+$ (**)

Questions:

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

2. How do I show (*)?

3. How do I show (**)?

**tonio** · Mar 9th 2010, 04:08 AM

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 $x<y<x^+$ (**)

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 $S:= \{s\in X\;;\;x\leq s\}$ , 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

**Mimi89** · Mar 9th 2010, 04:57 AM

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!

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