Sierpinski’s Theorem for Additive Functions
We say that is an additive function if
1. Prove that there exist additive functions which are discontinuous with or without the Darboux Property.
2. Prove that for every additive function there exist two functions which are additive, have the Darboux Property, and .
The second part is similar to Sierpinski’s Theorem which states that every real function can be written as the sum of two real functions with Darboux property.
(A function has the Darboux property if for every , is an interval.)
Proof: We know that every additive function is uniquely determined by its values over a basis of regarded as a vector space over . Consider such a basis. In the following, the vector spaces are always considered over .
1. (i) Examples of additive function without Darboux property Pick a basis and . Every real number can be expressed uniquely by
and we can define on the basis by for . This generates an additive function which is injective. Indeed if , then implies
and since is linearly independent, all coefficients are zero, and finaly .
Suppose has Darboux property. Then injectivity implies that is monotone which finally gets us to for some constant . But this contradicts and for . Therefore does not have the Darboux property.
It is possible to construct another example by simply choosing all for . This implies that takes only rational values, and therefore it does not have the Darboux property.
(ii) Example of additive function with Darboux property
We know that is uncountable and . We need a way to produce additive functions which vanish on a dense subset of without being continuous. This is because any noncontinuous function is not locally upper/lower blunded, which means that the image of any interval is dense in . To build an additive Darboux function we need it to take every real value in every interval, and in particular it takes the value zero in every interval. Furthermore, if is additive and is dense in then the function has the Darboux property. Indeed, for an arbitrary interval and a real number such that we know that is dense in so there exists such that where . Therefore . This proves that and since the interval was arbitrary, has the Darboux property.
Choose and look at which has
as a basis. Since is infinite, it has the same cardinal as (“removing” one element does not modify its cardinality) and therefore we can find an isomorphism .
Define now the function ,
which is additive and for every . Moreover, it is surjective (since is surjective) and for every interval we have (if there exists such that so for every ). Therefore has the Darboux property.
Another construction is possible starting from a basis which contains and defining . We choose on the rest of the basis such that is bijective. Furthermore, consider the equivalence relation . We find that is constant on every equivalence class of and furthermore, every interval contains one representative of every class (since all classes are dense in ). Therefore and has the Darboux property.
Note that if an additive function is locally bounded (above/below) then it is continuous. Therefore, a discontinuous additive function with Darboux property always satisfies for every interval .
2. Pick . As above, we have two bases of and given by
We can find a bijection and define (this is just a way to split into two parts of the same cardinality). Furthermore, we define the bijections and define on by
Because of the definitions of we find that .
We can define morphisms such that (just extend them from the definition on the basis) and these morphisms are surjective.
Define now by . is of course additive and surjective. Moreover since every is dense in , takes every real value in every interval . To prove this, choose , fix and by surjectivity of there exists such that . But is dense in so there exists such that . Therefore has the Darboux property (since for every interval ) and is additive.
Now let’s look at :
 : .
 : .
 : .
 : .
Therefore the additive functions and coincide on which proves that on .
(The last part was inspired from M. Megan, A.L. Sasu, B. Sasu Differential Calculus through Exercices and Problems (in Romanian).)

February 3, 2014 at 9:33 pmSierpinski’s Theorem for Additive Functions – Simplification  Beni Bogoşel's blog