Shawn

else we can choose M to be convex

is every subspace convex ?

ok thanks

oh ok

@Secret are you referring to me ?

*note

not that x is fixed

If M is closed subspace of Hilbert space H .let x belongs to H , then x+M is closed convex subset of H ?

Also Tychonoff theorem proves for product topology

@Rithaniel It's not that easy (to prove in one line), Tychonoff theorem there for help

@TheTerriblePuddle medicalsciences.stackexchange.com or wait for TedShifrin

It might be possible that arbitrary cover i am referring to does not have an element having each set as product ?

Also in box topology or in product topology or both ?

@Rithaniel I didn't got , can you elaborate little ?

Is uncountable product of compact spaces compact ?

If X*X is hausdoff then is X hausdorff ?

@MartinSleziak I am going to delete my questions this site is so rude

Thanks Martin

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

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

(c,d]

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

suppose take x belonging to A

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

*may not

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

Yes

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 ?

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

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

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

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

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

How can we say that ?

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

Hi

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

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 ?

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

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

Now after posted on Meta , Gerry said - If you ** knew** that it was from a contest that used to be active, you ought to have included that information in your original post. If you didn't know, well, now you have learned something.

am i a fool to ask it and cheat in contest which is already over ?

I provided complete proof that solution to question of that contest is already available

I think strict actions should be taken against people who ask contest related question

i mean people are unnecessary making false allegations

but i was ...

i can delete that comment

i agree

Thanks Martin , there are many good people like you and Jose Carlos etc from this community and due to you people only it is popular.

very important math.meta.stackexchange.com/questions/30495/…