 |
S.O.S. Mathematics CyberBoard
|
| View previous topic :: View next topic |
| Author |
Message |
Steve9
Senior Member
Joined: 22 Oct 2004
Posts: 74
|
 Posted: Thu, 30 Nov 2006 21:43:25 UTC Post subject: Separable space? |
 |
|
this was a question I was asked on an exam in my topology class:
Show  , where  , has a countable dense subset.
what I tried to do is to look at the open set that form a basis for . i.e. the finite intersection  where each  is distinct in [0,1] and  is open in  .
then picking closed subintervals  so that we may construct a finite sequence consisting of these  and points of  .
I next tried to construct a surjective map from the set of all these sequences onto (this is where I think I got lost or maybe it was when I read the question).
I wasn't able to construct this map, but would this be even a valid solution? |
|
| Back to top |
|
 |
gremlin
Member of the 'S.O.S. Math' Hall of Fame
Joined: 29 Feb 2004
Posts: 455
Location: middle of a dense grey fog
|
| Posted: Fri, 1 Dec 2006 15:13:53 UTC Post subject: |
|
|
whoops.  |
|
| Back to top |
|
|
Kungsman
Member of the 'S.O.S. Math' Hall of Fame
Joined: 04 Sep 2004
Posts: 3401
Location: Uppsala, Sweden
|
| Posted: Fri, 1 Dec 2006 18:32:13 UTC Post subject: Re: Separable space? |
|
|
| Steve9 wrote: | this was a question I was asked on an exam in my topology class:
Show , where , has a countable dense subset.
what I tried to do is to look at the open set that form a basis for . i.e. the finite intersection where each is distinct in [0,1] and is open in .
then picking closed subintervals so that we may construct a finite sequence consisting of these and points of .
I next tried to construct a surjective map from the set of all these sequences onto (this is where I think I got lost or maybe it was when I read the question).
I wasn't able to construct this map, but would this be even a valid solution? |
This is a special case of a more general theorem, which I guess is unknown to most students and would probably not have sufficed as an answer.
Now, one could (and perhaps should) view IR^I as the space of functions defined on the closed unit interval, taking values in IR. Let us construct a countable subclass of this class of functions and show it is dense.
Let  be a countable basis for [0,1]. I hope you agree there is such a basis (a simple consequence of Lindelöf's theorem, since separable metric spaces are also second countable, but in an explicit case like this you can think of open intervals having rational end-points, say).
So we consider the aggregate of functions
(the support here need not be closed; I only mean the function to be 0 on the complement of the B_{k_i}) which will be countable, being determined by a choice of the countable subfamily of B consisting of finitely many disjoint intervals together with a choice of a finite subset of  , which there are only countably many of.
To show  is dense in IR^I, let  be a basic open set in IR^I. By definition of the (product) topology on IR^I there is a finite set  such that  for  and for  U_x is some non-empty open subset of IR. For every x in J, choose elements from B so that these x ends up in different basic open sets, and also pick a rational number in U_x for all these x (in J).
The corresponding function in so determined belongs to U, so is dense in IR^I. Q.E.D. |
|
| Back to top |
|
|
|
|
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum
|
|
FAQ 
Search 
Memberlist 
Usergroups
Register
Profile 
Log in to check your private messages 
Log in 
