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Given the set $\mathbb{D}$ which contains all definable real numbers. The definition must not be infinite long. E.g. it contains $12$, $-3$, $\frac{1}{12}$, $\sqrt{2}$, $\pi^2$, $i+e$, Chaitin's constant and many other numbers.

$$ f(x)= \begin{cases} 1 &\quad\text{if }x\in\mathbb{D}\\ 0 &\quad \text{else} \end{cases} $$

Is it true that this function is periodic for every number which can be named / specified by anyone (which is equivalent to the set $\mathbb{D}$, or isn't it?)?

Thank you very much

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  • $\begingroup$ I define $U$ to be the first undefinable number ... $\endgroup$ – Joel Reyes Noche Jun 26 '16 at 0:44
  • $\begingroup$ @JoelReyesNoche: Then you need first to define a well-ordering of all the numbers ... $\endgroup$ – hmakholm left over Monica Jun 26 '16 at 0:48
  • $\begingroup$ @HenningMakholm, hmm, good point. $\endgroup$ – Joel Reyes Noche Jun 26 '16 at 0:49
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    $\begingroup$ Also, it is important to know that it is consistent with ZFC that every real number is definable in the language of set theory, so if that is what we mean by "definable" then we can't prove in ZFC that the function $f$ is nonconstant. $\endgroup$ – Carl Mummert Jun 26 '16 at 1:00
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    $\begingroup$ @user1952009: perhaps it is obvious, but as this post explains, the usual reasoning to show there are undefinable numbers isn't sound: mathoverflow.net/questions/44102/… $\endgroup$ – Carl Mummert Jun 26 '16 at 1:03
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There are some subtleties buried in the concept of "definable number". What exactly does it mean to you?

If you call a number "definable" if it satisfies a formula $\varphi(x)$ in the language of set theory such that $\forall x\forall y(\varphi(x)\land\varphi(y)\to x=y)$ is true, then you run into the problem that "definable" is not itself a property that can be expressed in set theory. So your $\mathbb D$ may not exist at all. (Or it may, in an appropriately small model of ZFC, equal $\mathbb C$ itself, such that your $f$ is actually constant).

On the other hand, you can speak about definability in some particular restricted language, such as $(\mathbb C,\mathbb R,\mathbb Z,0,1,+,\cdot)$. Then not everything you can describe using free-wheeling set theory will be definable, but the examples you give will, if the language is rich enough, which the one suggested above is. And in that case, then yes: Every definable number is a period for your function.

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  • $\begingroup$ Thanks for this good answer. Probably i should have written more detailed which definition of "definable number" i used. I only got information from this wikipedia article: en.wikipedia.org/wiki/… $\endgroup$ – Kevin Meier Jun 26 '16 at 9:11

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