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
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.