## Sierpinski’s Theorem for Additive Functions – Simplification

1. If is a solution of the Cauchy functional equation which is surjective, but not injective, then has the Darboux property.

2. For every solution of the Cauchy functional equation there exist two non-trivial solutions of the same equation, such that and have the Darboux property and .

These two results were proven in this post. The version presented here is a simplified one, identifying exactly what we need in order to obtain the desired results.

*Proof:* 1. First note that must be non-trivial. If would be trivial, then for some . The hypothesis surjective proves that , but then is injective. Contradiction.

Since is not injective, there exists such that , and using the properties of we have that for every .

Pick an interval and a value . Since is surjective, there exists such that . Since the set is dense in , there exists such that . Therefore and . This proves that . Since this happens for every interval , it follows that has the Darboux property.

2. Consider a basis of considered as a vector space over and . Since there exists a bijection . Denote and . Therefore, we have just partitioned into two uncountable sets and . This allows us to construct another two bijections and .

Now we can define the two functions with the needed properties. As we have seen in the introductory part of this section, a solution of the Cauchy functional equation is well and uniquely defined if we know the values of the solution on a basis of over , which in our case is .

We define and by their values over as follows:

and

The two functions constructed above, are solutions to the Cauchy functional equation. Moreover, it is easy to see that for every , which implies .

Since is a bijection from to it follows that is surjective, and because we see that is not injective. Using 1. we conclude that is non-trivial and has the Darboux property. By a similar argument is non-trivial and has the Darboux property.