Well Orders and Total Orders .... Searcoid Definition 1.3.10 ....

[Math Amateur]

I am reading Micheal Searcoid's book: "Elements of Abstract Analysis" ... ...

I am currently focused on understanding Chapter 1: Sets ... and in particular Section 1.3 Ordered Sets ...

I need some help in fully understanding some remarks by Searcoid following Definition 1.3.10 ...

Definition 1.3.10 and the remarks following read as follows:
View attachment 8427
View attachment 8428In Searcoid's remarks following Definition 1.3.10 we read the following ...

"... ... every well ordered set is totally ordered ... ... Can someone please help me to prove that every well ordered set is totally ordered ... ...Help will be appreciated ...

Peter

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It may help MHB memebers reading the above post to have access to Definition 1.3.4 ... so I am providing access to the same ... as follows ...
View attachment 8429
Hope that helps ...

Peter

 

[castor28]

Hi Peter,

You just have to apply the definition to subsets of two elements.

More explicitly, if $S$ is well ordered and $\{x,y\}\subset S$ with $x\ne y$, then $\{x,y\}$ has a smallest element. If that element is $x$, then $x<y$; if that element is $y$, then $y<x$.

 

[Joppy]

Hi Peter,

You just have to apply the definition to subsets of two elements.

More explicitly, if $S$ is well ordered and $\{x,y\}\subset S$ with $x\ne y$, then $\{x,y\}$ has a smallest element. If that element is $x$, then $x<y$; if that element is $y$, then $y<x$.

Sorry to butt in.. But what if $S$ is singleton? They also have a minimum value since no number is greater than itself (...). How do we compare it to a totally ordered set which (seems to) require at least two elements?

edit: i.e. if it doesn't make sense consider a totally ordered set which is singleton, how can we say all well ordered sets are totally ordered? (I'm assuming we can't and that I've missed a definition or am thinking something really silly)..

 

[castor28]

Sorry to butt in.. But what if $S$ is singleton? They also have a minimum value since no number is greater than itself (...). How do we compare it to a totally ordered set which (seems to) require at least two elements?

edit: i.e. if it doesn't make sense consider a totally ordered set which is singleton, how can we say all well ordered sets are totally ordered? (I'm assuming we can't and that I've missed a definition or am thinking something really silly)..

Hi Joppy,

A singleton set is totally ordered, because the definition is vacuously true.

The definition of a totally ordered set $S$ is equivalent to: if $\{x,y\}\subset S$ and $x\ne y$, then $x<y$ or $y<x$.

If $S$ is a singleton, the condition $x\ne y$ is always false. As the antecedent of the implication is false, the implication itself (the definition) is true.

Note that it is quite common to define an order relation as a reflexive, antisymmetric and transitive relation (like $\le$), and to modify the definitions accordingly. In that case, things are a little simpler.

 

[steenis]

https://math.stackexchange.com/questions/463331/well-ordered-and-totally-ordered-setLook at the third answer .

 

[Math Amateur]

https://math.stackexchange.com/questions/463331/well-ordered-and-totally-ordered-setLook at the third answer .

Thanks to Castor28, Joppy and Steenis for their help with the above issue ...

Peter