Is well ordered set countable , if yes then we can prove that every element must have predecessor (except smallest and largest )

@Shawn No, my example is countable and not every element has a predecessor.

I am reading a proof from munkres of Theorem :: Every well ordered set X is normal in the order topology .In the proof it is given "let A and B be disjoint closed sets in X and they don't contain smallest element .Then for each $a∈ A$ , there exists a basis element about a disjoint from $B $ , it contains some interval of the form $(x,a]$ ".How do i justify this ?

@Shawn doesn't letting x be the smallest element of $A$ work?

"there exists a basis element about a disjoint from B ; it contains some interval " , there is semicolon not comma .Sorry !

No taking x be the smallest element does not work since B can be subset of (x,a]

@Shawn Oh I see, yes this is annoying.You really need to used that A and B are closed.

Hi

Can you tell me how to prove that part ? @LeviRyffel

@Shawn Hey. The complement $B^c$ of $B$ is open, so around any point in $B^c$ (in particular $a$) there exists an interval in $B^c$. Does this help?

How can we say that ?

I mean around every point in B^c (B complement) there exist an interval , how can we prove that ?

@Shawn Because the intervals form a basis

But there might be a point in B complement such that all intervals containing it also intersect B !

No because the complement of B is open

Ok , hence you are saying that atleast B^c (B complement) will one such open set that will contain that point ?

But how to claim that open set will be of form (x,a] ?

Can you just open Munkres once and read that proof , you will get what i am trying to say

Also one more doubt if in a well ordered set there exists an element say x such that it has predecessor then is it true that all element of set will have predecessor ?

Sorry didn't want to do that

Yes

your doubt is wrong, as my example shows. Do you know what $\omega$ is?

Every element in $\omega + 1$ that is not equal to $\omega$ has a predecessor

No , but i got some idea from your answer like immediate predecessor does not exist

*may not

$\omega$ is basically the natural numbers with the usual order, and $\omega + 1$ is the natural numbers together with infinity. Maybe this is easier for you to think about.

Do you want a proof why $(x, a]$ is open?

No i know that is correct due to well order definition . Here i will explain something and just tell if i am correct , for each a belonging to A , there exist a basis element disjoint from B because

suppose take x belonging to A

then since B^c is open , hence it must be union of open intervals of the form

(c,d]

Thus x must be inside such interval (c,d] which is disjoint from B !

yes

Thanks , that was very silly doubt .Thanks for your help and have a nice day !

@Shawn No problem, it was fun to think about this :)