Author 
Message 
lataianu bogdan science forum beginner
Joined: 25 Sep 2005
Posts: 39

Posted: Mon Jul 17, 2006 7:11 pm Post subject:
In which (ordered) spaces an increasing sequence has a limit?



Suppose we have a Haussdorf topological space.
When an increasing sequence has a limit( finite or infinite)?
For example: Let NxN with order topology. The order is the lexicographic order.
I think that NxN has this property. For example the sequence:(0,1), (0,2), ... , (0,n), ... has limit (1,0) since in any neighbourhood of (1,0) there will be an infinity of elements of the sequence.
The same property doesn't hold in RxR.
Does this happen because NxN is wellordered with order topology?
Thank you. 

Back to top 


Robert B. Israel science forum Guru
Joined: 24 Mar 2005
Posts: 2151

Posted: Mon Jul 17, 2006 9:34 pm Post subject:
Re: In which (ordered) spaces an increasing sequence has a limit?



In article <10901063.1153163564343.JavaMail.jakarta@nitrogen.mathforum.org>,
lataianu bogdan <lataianu(withnobrackets)@math.usask.ca> wrote:
Quote:  Suppose we have a Haussdorf topological space.
When an increasing sequence has a limit( finite or infinite)?

It has to be a partially ordered set in the first place, to have
"increasing" make any sense.
Quote:  For example: Let NxN with order topology. The order is the lexicographic order.
I think that NxN has this property.

What about the sequence (0,0), (1,1), (2,2), ...?
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada 

Back to top 


William Elliot science forum Guru
Joined: 24 Mar 2005
Posts: 1906

Posted: Tue Jul 18, 2006 5:04 am Post subject:
Re: In which (ordered) spaces an increasing sequence has a limit?



On Mon, 17 Jul 2006, lataianu bogdan wrote:
Quote:  Suppose we have a Haussdorf topological space.
When an increasing sequence has a limit( finite or infinite)?

Suppose the space has an order.
Quote:  For example: Let NxN with order topology. The order is the lexicographic
order. I think that NxN has this property. For example the
sequence:(0,1), (0,2), ... , (0,n), ... has limit (1,0) since in any
neighbourhood of (1,0) there will be an infinity of elements of the
sequence.
The same property doesn't hold in RxR.
Does this happen because NxN is wellordered with order topology?

No. Look at lex [0,1)x[0,1) and lex [0,1]x[0,1] 

Back to top 


William Elliot science forum Guru
Joined: 24 Mar 2005
Posts: 1906

Posted: Tue Jul 18, 2006 5:05 am Post subject:
Re: In which (ordered) spaces an increasing sequence has a limit?



On Mon, 17 Jul 2006, Robert Israel wrote:
Quote:  lataianu bogdan <lataianu(withnobrackets)@math.usask.ca> wrote:
Suppose we have a Haussdorf topological space.
When an increasing sequence has a limit( finite or infinite)?
It has to be a partially ordered set in the first place, to have
"increasing" make any sense.
For example: Let NxN with order topology. The order is the
lexicographic order. I think that NxN has this property.
What about the sequence (0,0), (1,1), (2,2), ...?
It has an infinite limit? 


Back to top 


Robert B. Israel science forum Guru
Joined: 24 Mar 2005
Posts: 2151

Posted: Tue Jul 18, 2006 7:23 pm Post subject:
Re: In which (ordered) spaces an increasing sequence has a limit?



William Elliot wrote:
Quote:  On Mon, 17 Jul 2006, Robert Israel wrote:
lataianu bogdan <lataianu(withnobrackets)@math.usask.ca> wrote:
Suppose we have a Haussdorf topological space.
When an increasing sequence has a limit( finite or infinite)?
It has to be a partially ordered set in the first place, to have
"increasing" make any sense.
For example: Let NxN with order topology. The order is the
lexicographic order. I think that NxN has this property.
What about the sequence (0,0), (1,1), (2,2), ...?
It has an infinite limit?

Hmm. I must have missed the "(finite or infinite)" somehow.
OK, suppose X is a totally ordered set with the order topology. To
allow "infinite limits", you can adjoin a new element "+infinity"
which is greater than all the original members of X.
A totally ordered set is said to be "complete" if every nonempty
subset with an upper bound has a least upper bound. What we're
dealing with here, I guess, would be called "sequential
completeness". Of course, the least upper bound of an increasing
sequence is the limit of that sequence.
Exercise: If X is separable (i.e. it has a countable subset D that is
dense in the order topology) and every increasing sequence has
a limit, then X is complete.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada 

Back to top 


William Elliot science forum Guru
Joined: 24 Mar 2005
Posts: 1906

Posted: Wed Jul 19, 2006 10:43 am Post subject:
Re: In which (ordered) spaces an increasing sequence has a limit?



From: Robert Israel <israel@math.ubc.ca>
Newsgroups: sci.math
Subject: Re: In which (ordered) spaces an increasing sequence has a limit?
William Elliot wrote:
Quote:  On Mon, 17 Jul 2006, Robert Israel wrote:
lataianu bogdan <lataianu(withnobrackets)@math.usask.ca> wrote:
Suppose we have a Haussdorf topological space.
When an increasing sequence has a limit( finite or infinite)?
It has to be a partially ordered set in the first place,
For example: Let NxN with order topology. The order is the
lexicographic order. I think that NxN has this property.
What about the sequence (0,0), (1,1), (2,2), ...?
It has an infinite limit?
Hmm. I must have missed the "(finite or infinite)" somehow.
OK, suppose X is a totally ordered set with the order topology. To
allow "infinite limits", you can adjoin a new element "+infinity"
which is greater than all the original members of X.
A totally ordered set is said to be "complete" if every nonempty
subset with an upper bound has a least upper bound. What we're
dealing with here, I guess, would be called "sequential
completeness". Of course, the least upper bound of an increasing
sequence is the limit of that sequence.

Ascending sequence completeness.
Quote:  Exercise: If X is separable (i.e. it has a countable subset D that
is dense in the order topology) and every increasing sequence has
a limit, then X is complete.

Hm, R doesn't fulfill the premise.
Indeed X must have a maximum element.
Assume X unbounded without top. List D = { d1, d2, ... }
Pick x1 in (d1,oo), x2 in (max(x1,d2), oo)
... x_n in (max(x_(n1), d_n), oo) ...
Let x be the limit of (x_j)_j. D <= x, ie x upper bound D.
If some a > x, then some d in D /\ (x,oo); x < d, no!
Thus x is maximum element of X.
Tho (0,1) doesn't fulfill the premise, (0,1] and [0,1] do.
Assume nonnul A subset X.
down A = { x  some a in A with x <= a }
case down A has maximum element a: a = sup down A = sup A
case down A doesn't have maximum element:
D /\ down A dense subset down A
From above down A has some ascending sequence (a_j)_j
without a limit (in down A)
(a_j)_j > a in X\down A
down A <= a, ie a is upper bound of down A
if down A <= x, then a <= x.
a = sup down A = sup A
QED. (hopefully)
Exercise: sup down A = sup A
The difference between 'sequential completeness' and 'ascending sequence
completeness' is the first connotates order symmetry or order duality
while the later is definely without such property.
You have you seen the notation (down arrow)A for down A where
(down arrow) is a single graphic character same size as A?
 

Back to top 


Google


Back to top 



The time now is Fri Apr 26, 2019 2:09 pm  All times are GMT

