Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In the wikipedia article about descriptive set theory I read that $\mathbb{R}$ (with its usual topology) is a Polish space, and that every Polish space

1) can be obtained as a continuous image of the Baire space $\mathcal{N}$

2) can be obtained as the image of a continuous bijection defined on a closed subset of the Baire space.

Then I learn from the wikipedia article on Baire space that it is actually homeomorphic to the irrational numbers with their usual subspace topology iherited from the real line.

So my questions:

Can we describe explicitely a surjective continuous map from the irrationals to $\mathbb{R}$?

Same for a continuous bijection from a closed subset of the irrationals to $\mathbb{R}$.

share|improve this question
3  
(Perhaps I should have posted this on MathSE instead of MO?) –  Qfwfq Nov 11 '12 at 23:24
add comment

6 Answers

up vote 15 down vote accepted

For any irrational number $x$, let $f(x)$ be the real number arising from the integer part of $x$, together with every other digit of the rest of the expansion of $x$.

This is surjective, since one may interleave the digits of any real $y$ with any nonrepeating pattern, and thereby find an irrational $x$ with $f(x)=y$. This is continuous, since if $x_n\to x$, then $f(x_n)\to f(x)$.

share|improve this answer
1  
Surely I'm being dense, but what do you mean by "taking every other digit of the expansion of $x$"? –  Qfwfq Nov 11 '12 at 23:44
1  
I mean, for example, that $3.1415926535897932384\ldots$ maps to $3.1196387334\ldots$ –  Joel David Hamkins Nov 11 '12 at 23:47
    
Ok, so you meant: the integer part is the same while the $n$-th digit of the fractionary expansion is replaced by the $(2n-1)$-th digit. –  Qfwfq Nov 12 '12 at 14:15
    
(or: the $n$-th digit is replaced by the $\sigma(n)$-th, where $\sigma$ is any monotone function) –  Qfwfq Nov 12 '12 at 14:17
1  
Yes, that's right. But note the slight complication that the resulting expansion may be something like $0.999999\ldots$, which means it is really $1$. –  Joel David Hamkins Nov 12 '12 at 14:26
show 1 more comment

Let $C$ be a space-filling curve, i.e. a continuous function from $\mathbb R$ onto ${\mathbb R}^2$. Then the first component $C(t)_1$ is a continuous function from $\mathbb R$ onto $\mathbb R$ that hits each real number uncountably many times. Since there are only countably many rationals, the restriction to the irrationals is also surjective.

share|improve this answer
    
Thank you for the very clear answer to my first question. What about the second question on continuous bijection from a closed set of irrationals? –  Qfwfq Nov 12 '12 at 0:05
add comment

For the second question, remember that the subset only has to be closed in the irrationals, not as a subset of $\mathbb{R}$. With that in mind, you can take the closed subset to be the irrationals in $(-1,1)$ union some countable discrete set of irrationals. On $(-1,0)$, define $f$ by $x\mapsto \frac{1}{x}+1$; on $(0,1)$, by $x\mapsto \frac{1}{x}-1$, and on the discrete set, choose some bijection with $\mathbb{Q}$.

share|improve this answer
add comment

A surjection from $\mathbb{N}^\mathbb{N}$ to $\mathbb{R}$ is given by $$f \mapsto f(0) - f(1) + \sum_{k = 2}^\infty \frac{\min(f(2k),1)-\min(f(2k+1),1)}{2^{2k-4}}$$ A homeomorphism from the irrational numbers (with the Euclidean topology) to $\mathbb{N}^\mathbb{N}$ is given by the continued-fraction expansion.

The displayed map is onto, constructively. For the continued-fraction expansion map to be constructive, we need to define irrational numbers as those reals $x \in \mathbb{R}$ for which $|x - q| > 0$ for all $q \in \mathbb{Q}$ (as opposed to those reals which are not rational).

share|improve this answer
    
The only way I can make heads or tails of this formula is that $f$ is not an irrational number as requested by the OP, but an element of the Baire space, $f\colon\mathbb N\to\mathbb N$. Am I wrong? –  Emil Jeřábek Nov 12 '12 at 16:31
    
@Emil Ah! It now makes sense. –  Andres Caicedo Nov 12 '12 at 17:00
    
Oops, I added a comment to that effect. –  Andrej Bauer Nov 12 '12 at 17:23
add comment

Here's another simple explicit function that maps the irrationals onto the real numbers.

Consider the continuous piece-wise linear function $f:\mathbb{R}\rightarrow\mathbb{R}$, with slope $f'(x)=-1$ or $f'(x)=+1$ according whether $\lfloor x/\sqrt 2\rfloor=0\, \mathrm{ mod} 3 $, or not (like this). So $f(x)=x/3+O(1)$ as $|x|\to\infty$ and it is therefore surjective; precisely, any of its fiber has three points, and the arithmetic mean of some two of these is an odd multiple of $\sqrt{2}$. This means that the equation $f(x)=c$ has always an irrational solution, whatever is $c\in\mathbb{R}$.

share|improve this answer
add comment

Count the rational numbers as $(a_1,a_2,a_3,\ldots)$. Let $f:\mathbb R\to \mathbb R$ be defined at the nonpositive integers by $f(n)=(-1)^nn$, and on the positive integer multiples of $\sqrt 2$ by $f(n\sqrt 2)=a_n$. Let $f$ be piecewise linear between the points just defined. Then the restriction of $f$ to $\mathbb R\setminus \mathbb Q$ is a continuous surjection from the irrationals to $\mathbb R$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.