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Cauchy's Functional Equation Revisited

It is commonly known that the solution for Cauchy's Functional Equation

\[ f:\mathbb{Q} \rightarrow \mathbb{Q}, \quad f(x)+f(y)=f(x+y)\]

has solutions \(f(x)=cx\). (If you don't know how to prove this, I ask for you to attempt to prove this)

But now, if we instead change this function from being in the rationals to being the reals, we get a strange property.

Let \(f\) be a function \(f:\mathbb{R} \to \mathbb{R}\) such that

\[\begin{align} f(x) + f(y) &= f(x+y) \forall x, y\\ f(1) &= 1\\ \text{ The function is not }f(x) & = x \end{align}\]

Consider the graph of \( y = f(x) \) on the plane. Prove that for any disc, no matter how small, there always exists a point that lies on the graph.

For this proof, you must accept the Axiom of Choice as true.


This question came from a corollary I used in APMO 2016 Q5.

Note by Sharky Kesa
12 months ago

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Wikipedia :P Deeparaj Bhat · 12 months ago

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@Deeparaj Bhat How do you know Wikipedia is correct? Anyone can edit it. Sharky Kesa · 12 months ago

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@Sharky Kesa Anyone can edit wikipedia :v

For math homework, i usually go here if i'm confused

Because you guys are more trusted than wikipedia Jason Chrysoprase · 12 months ago

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@Sharky Kesa I've seen their proof. It was right in my opinion. Deeparaj Bhat · 12 months ago

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@Calvin Lin I'm saying on the graph of \((x, f(x)\). Sharky Kesa · 11 months, 4 weeks ago

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@Sharky Kesa K, edited. Makes sense now. Calvin Lin Staff · 11 months, 4 weeks ago

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