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October 28th, 2009, 08:32 PM  #1 
Newbie Joined: Oct 2009 Posts: 2 Thanks: 0  A Continuous surjective function: irrationals > rationals Question is: Show that there is a continuous surjective function f : [0, 1]\Q > [0, 1] ^ Q (^ is intersection) This makes sense since rationals and irrationals are both "dense" in the interval. But "this makes sense" is not a proof. Please help me! 
October 29th, 2009, 01:55 AM  #2 
Senior Member Joined: Jan 2009 From: Japan Posts: 192 Thanks: 0  Re: A Continuous surjective function: irrationals > rationals
As an example surjective function: Let f(a) = b if b is (a + pi  3) or (a + pi  4) (whichever lies in [0, 1]). It should be clear that this is surjective. Making it continuous depends on the topology  I'm a fan of the discrete topology if you get to choose. 
October 29th, 2009, 12:39 PM  #3  
Newbie Joined: Oct 2009 Posts: 2 Thanks: 0  Re: A Continuous surjective function: irrationals > rationals Quote:
But all irrationals in the interval have to be mapped to certain rationals, otherwise the function is not welldefined. Not all irrationals can be expressed in the form of (a + pi  3) or (a + pi  4) , where a is a rational. So to which rationals are those irrationals (which cannot be expressed that way) mapped onto?  
October 29th, 2009, 04:10 PM  #4 
Senior Member Joined: Jan 2009 From: Japan Posts: 192 Thanks: 0  Re: A Continuous surjective function: irrationals > rationals
Yes, that's why I say making it continuous depends on the topology. I don't see a function off the top of my head that is continuous in the standard metric; someone else might.

October 31st, 2009, 05:40 PM  #5 
Member Joined: Oct 2009 Posts: 64 Thanks: 0  Re: A Continuous surjective function: irrationals > rationals
Well, the irrationals have a greater cardinality than the rationals, which poses problems if you want a 'normal' function: one that is differentiable except at a finite set of points on any finite interval.


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