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Of course, no continuous real valued non-constant function can attain only rational or irrational values, but can there be a pair of nowhere-constant continuous functions f and g such that for all x, at least one of f(x) and g(x) is rational? Or maybe a countable collection of continuous functions, {f1, f2...} such that for all x there is n such that fn(x) is rational?

Thanks

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    $\begingroup$ I assume that you want f(x) and g(x) to be nonconstant? $\endgroup$
    – Andy Putman
    Commented Sep 10, 2010 at 19:45
  • $\begingroup$ Yeah - edited to fix that. $\endgroup$
    – mathahada
    Commented Sep 10, 2010 at 19:47
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    $\begingroup$ You probably want non-locally constant, since there are trivial counterexamples with two functions making steps with constant rational value on intervals. $\endgroup$
    – Joel David Hamkins
    Commented Sep 10, 2010 at 19:48
  • $\begingroup$ R is connected, so locally constant is equivalent to constant. $\endgroup$
    – user5810
    Commented Sep 10, 2010 at 19:50
  • $\begingroup$ Sorry, I meant non-constant on any interval. $\endgroup$
    – Joel David Hamkins
    Commented Sep 10, 2010 at 19:59

3 Answers 3

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If you allow the functions to be constant on some intervals, then there are some easy examples, and Ricky has provided one.

But if you rule that out, then there can be no examples, even with countably many functions. To see this, suppose that $f_n$ is a list of countably many continuous functions which are never constant on an interval. Enumerate the pairs $(r,n)$ of rational numbers $r$ and natural numbers $n$ in a countable list $\langle (r_0,n_0), (r_1,n_1),\ldots\rangle$. Let $C_0$ be any closed interval. If the closed interval $C_i$ is defined, consider the function $f_{n_i}$ and the rational value $r_i$. Since $f_{n_i}$ is not constant value $r_i$ on $C_i$, we may shrink the interval to $C_{i+1}\subset C_i$ such that $f_{n_i}$ on $C_{i+1}$ is bounded away from $r_i$. By compactness, there is some $x\in C_i$ for all $i$. Thus, $f_n(x)$ is not $r$ for any rational number $r$.

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    $\begingroup$ And this argument amounts to the proof of the Baire Category theorem, mentioned by Mikhail... $\endgroup$
    – Joel David Hamkins
    Commented Sep 10, 2010 at 20:21
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Since your functions are locally non-constant, the preimage of any (rational) point is nowhere dense (in $\mathbb{R}$; if a preimage of a point with respect to a continious function is dense on an interval, then the function is constant on this interval). Hence the union of all preimages of all rational points is of Baire category one (in $\mathbb{R}$); so it is not equal to the whole $\mathbb{R}$.

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    $\begingroup$ +1. This way of thinking about it shows that it is consistent with $ZFC+\neg CH$ that you can't even do it with $\aleph_1$ many functions (or more, even), since it is known to be consistent with $\neg CH$ that the ideal of meager sets has more than countable additivity. $\endgroup$
    – Joel David Hamkins
    Commented Sep 10, 2010 at 20:44
  • $\begingroup$ +1. @Joel, do you know where I could find a proof of that consistency result? (I suppose what I'm really interested in is a respectable article stating it, since I probably wouldn't understand a proof.) $\endgroup$
    – user5810
    Commented Sep 10, 2010 at 20:51
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    $\begingroup$ It is part of Cichon's diagram (see en.wikipedia.org/wiki/Cichon_diagram). This seems to be one of the standard results in the field of cardinal characteristics. It is probably in the survey article by Andreas Blass: math.lsa.umich.edu/~ablass/set.html $\endgroup$
    – Joel David Hamkins
    Commented Sep 10, 2010 at 20:58
  • $\begingroup$ Hey Joel. Why it doesn't work with uncountably many functions? It is enough to have f(x) = ax for any real a. $\endgroup$
    – mathahada
    Commented Sep 10, 2010 at 21:44
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    $\begingroup$ I am saying merely that it is consistent with the axioms of set theory that you could extend the result to $\aleph_1$ many functions, but of course, this is only possible when $\aleph_1\lt 2^{\aleph_0}$, which is to say, when the Continuum Hypothesis fails. The result can never hold for continuum $2^{\aleph_0}$ many functions, as your example shows. $\endgroup$
    – Joel David Hamkins
    Commented Sep 10, 2010 at 21:55
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f(x) := closest point in [-2,-1] to x
g(x) := closest point in [+1,+2] to x

If x≤0, then g(x) is rational.
If 0≤x, then f(x) is rational.

The question becomes more interesting if you demand that the functions be nowhere locally constant.

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  • $\begingroup$ I was not familiar with this term up until now but yes, that's what I meant. Perhaps a better formulation would be: is there a curve in the plane such that all of its points have at least one rational coordinate $\endgroup$
    – mathahada
    Commented Sep 10, 2010 at 19:54

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