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Please help me find an injection, $F$, from the set of real numbers into itself such that $F(x) - F(y)$ is an irrational number for any two distinct real numbers $x$ and $y$.

Thank you.

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This is impossible: If $x = y$, you have $F(x) - F(y) = F(x) - F(x) = 0$, which is rational. Are there any additonal conditions? –  Johannes Kloos May 16 '12 at 15:07
    
Do you mean that $F(x)-F(y) \notin \mathbb{Q}$ for all distinct pairs of real numbers $x$ and $y$? –  mjqxxxx May 16 '12 at 15:12
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Hint: use the axiom of choice to select one element from each set of the form $x+\mathbb Q$. You'll have to find support for the claim that this gives you a set of cardinality continuum. –  user31373 May 16 '12 at 15:18
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If the range of $F$ is measurable with positive measure, then the set of differences contains an interval about zero, so contains some rationals. So for your example you need either something non-measurable (as Leonid did) or measure zero. For the latter, think of some things like Cantor sets. –  GEdgar May 16 '12 at 15:26
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What do you mean by "find"? You can prove that such an $F$ must exist, in ZFC, but the word "find" might mean you want a definition. Given that it ultimately involves the axiom of choice, there is probably no "constructive" definition. –  Thomas Andrews May 16 '12 at 15:41

3 Answers 3

Consider $\mathbb R$ as vector space over $\mathbb Q$. Complete the set $\{1\}$ to a basis, call it $B$. There is a bijection $f \colon B \to B - \{1\}$ which extends to a linear transformation $F \colon \mathbb R \to \mathbb R$ that satisfies the requirement, because it is irrational everywhere except 0.


A constructive answer without AC:

Lemma:

There is an absolutely convergent sequence of real numbers $x_i$ such that $\sum a_i x_i$ is irrational for every $a_i \in \{-1,0,1\}$ except all $a_i = 0$.

Proof assuming the lemma:

Since $\mathbb R$ and $\{0,1\}^{\mathbb N}$ have equal cardinality, it is enough to show $f \colon \{0,1\}^{\mathbb N} \to \mathbb R$.

Define $f(a_1 a_2 \dots) = \sum a_i x_i$. The difference $f(x)-f(y)$ is always irrational by the lemma.

Proof of the lemma:

Parition $\mathbb N$ into infinitely many infinite subsets $A_i$ and let $x_i = \sum_{k \in A_i} 10^{-k!}$. The series $\sum a_i x_i$ is, except when all $a_i = 0$, a Liouville number and therefore irrational.

(Alternatively, the lemma can be shown with precise but boring diagonalization.)

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Very cool! I thought this result actually required choice, but I can't spot any hidden uses in your proof at least at first glance; I could easily be wrong... –  Steven Stadnicki Jun 23 '12 at 7:27

Let us pull this question off from unanwered questions' hell, following Leonid's idea. First, taking additive groups, we look at the quotient $\,\mathbb R/\mathbb Q\,$ . By Lagrange's theorem, $$|x+\mathbb Q|=|\mathbb Q|=\aleph_0\,\,,\,\,\forall x\in \mathbb R,$$and since the different cosets are in fact a partition of the set $\,\mathbb R\,$ , their (disjoint, of course) union is $\,\mathbb R\,$ . It follows that there are uncountable many different cosets as above (why?).

Now, we can make use of the Axiom of Choice in its form called Zermelo's Well Ordering Theorem, so we can introduce a well ordering in $\,\mathbb R\,$ which in turn induces a well ordering in $\,\mathbb R/\mathbb Q\,$, and then form any index (out of an uncoutable set) $\,\alpha\,$ , we cand define $$f:\mathbb R\longrightarrow \mathbb R\,\,,\,\,f(x_\alpha):=y_\alpha$$where $\,y_\alpha\,$ belongs to the $\,\alpha$-th coset in $\,\mathbb R/\mathbb Q\,$ (also here we use AC...)

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I think it's not quite enough to show the number of cosets is uncountable; you either have to assume the Continuum Hypothesis, or you have to show that the cosets are in one-one correspondence with the reals. –  Gerry Myerson Jun 23 '12 at 0:28
    
Ok, fair enough...I think there exists somewhere a proof that $\,\mathbb R/\mathbb Q=2^{\aleph_0}\,$, but if nothing better can be gotten then I'd be willing to accept the C.H...:) –  DonAntonio Jun 23 '12 at 0:58
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Certainly $|\mathbb{R}/\mathbb{Q}|=2^{\aleph_0}$ assuming AC, since $|\mathbb{R}/\mathbb{Q}|\aleph_0=2^{\aleph_0}$, and with AC, the product of two infinite cardinals is their maximum. –  Harry Altman Jun 23 '12 at 7:31

Under AC there is a well-order of $\mathbb R$ with each element having $\lt \mathfrak c$ predecessors. So just go along the well-order assigning an allowable value to each real. Each previous element excludes $(\lt \mathfrak c) \times \aleph_0$ values, so you will have at least one available.

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