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I read somewhere that the proof depends on AC.

But how about showing it explicitly i.e finding an explicit example that satisfies this?

I would like to see both proofs, the one that depends on AC and one that not (if it's possible).

Thanks! N.B I tried searching for it in google, but with no success.

MathematicalPhysicist
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  • @AsafKaragila check the old forum post here: http://www.tapuz.co.il/forums/viewmsg/457/12834712/%D7%9C%D7%99%D7%9E%D7%95%D7%93%D7%99%D7%9D_%D7%95%D7%94%D7%A9%D7%9B%D7%9C%D7%94_%D7%92%D7%91%D7%95%D7%94%D7%94/%D7%9E%D7%AA%D7%9E%D7%98%D7%99%D7%A7%D7%94 – MathematicalPhysicist Apr 26 '20 at 21:02

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(This is based on this blogpost, which I found through this answer.)

Pick a Hamel basis for $\Bbb R$ over $\Bbb Q$, that is a set of reals, $B$, such that every real number can be written as a unique linear combination of elements of $B$, with rational coefficients.

For $r\in B$ and $x\in\Bbb R$, let $x_r$ be the rational coefficient of $r$ in the linear combination resulting in $x$. Note that for all but finitely many $r$'s, it will be $0$. So in other words, $x=\sum_{r\in B}x_rr$ for every real number.

Claim. For a fixed $r$, the function $x\mapsto x_r r$ is periodic.

Proof. Note that if $r'\in B$ and $r'\neq r$, then $(x+r')_r=x_r$. $\square$

This easily generalises: if $A\subsetneq B$, then $x_A=\sum_{r\in A}x_r r$ is a well-defined function, and if $r\in B\setminus A$, then $x\mapsto x_A$ is periodic with period $r$.

Finally, we are ready to prove the claim about the identity function!

Fix $r\in B$, and let $f_1(x)=x_r r$, and $f_2(x)=x_{B\setminus\{r\}}$. Then $f_1+f_2=\operatorname{id}$, and by the above, both are periodic.

The axiom of choice was needed for obtaining the Hamel basis. Indeed, it is consistent without the axiom of choice that there is no such basis, in which case the above proof certainly does not work.

Of course, this is not a proof that the axiom of choice is required. One way to prove this is to show that any pair of periodic functions which can be added to the identity must be $\Bbb Q$-linearity, which under some conditions imply continuity, which means it is impossible. But I am not familiar with the theory of periodic functions enough to give you an answer off the cuff.

Arnoud Nijon
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Asaf Karagila
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  • So if I assume $\not AC$ then I get that there are no such functions? – MathematicalPhysicist Apr 12 '20 at 08:44
  • Assuming that choice fails tells you nothing. Just like saying that a set is finite tells you nothing about the size of the set, or what kind of elements are in that set. What I said is that maybe it is possible to prove that it is possible that with additional axioms, which contradict the axiom of choice (but not the rest of ZF), that maybe in these cases there is no such decomposition. – Asaf Karagila Apr 12 '20 at 08:48
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    The task of proving that this proposition depends on the axiom of choice is very difficult : you have to exhibit some model of ZF(without choice) where $f$ can't be written as the sum of two periodic functions.

    The proof given by Asaf above relies on AC. But to show that one can't find a proof of our proposition that doesn't use AC is another story.

     – Olivier Roche Apr 12 '20 at 09:05
  • @OlivierRoche so here's an interesting open problem to dwell on. :-) – MathematicalPhysicist Apr 12 '20 at 09:09
  • @AsafKaragila I'm too lazy to compute a forcing doing this, but I would enjoy a sketch of a proof. – Olivier Roche Apr 12 '20 at 09:10
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    @Olivier: That sounds like fun. – Asaf Karagila Apr 12 '20 at 09:14
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    @Olivier: I'm not sure how a forcing proof would look like. You maybe want to mimic the "nowhere differentiable continuous function", where we force with rational-based approximations, and then use absoluteness. But In that case we can rely on these rational-based approximation to be a ccc forcing; and we can use absoluteness due to continuity. In this case we have to approximate everywhere, and it's not obvious to me that this is a $\Pi^1_3$ statement, as the functions are not remotely continuous (in principle). So we're left with other type of approximations, but this is tantamount to Zorn. – Asaf Karagila Apr 12 '20 at 10:53