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October 28th, 2009, 08:32 PM   #1
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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!
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October 29th, 2009, 01:55 AM   #2
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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.
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October 29th, 2009, 12:39 PM   #3
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Re: A Continuous surjective function: irrationals --> rationals

Quote:
Originally Posted by cmusick
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.

But all irrationals in the interval have to be mapped to certain rationals, otherwise the function is not well-defined. 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?
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October 29th, 2009, 04:10 PM   #4
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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.
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October 31st, 2009, 05:40 PM   #5
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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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