If $I$ is an interval of real numbers , then is it true that any function $f:I \to \mathbb R$ can be written as $f=f_1+f_2$ , where $f_1 , f_2 : I \to \mathbb R$ have the Intermediate value property?
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See also this question: http://math.stackexchange.com/questions/100096/references-about-sierpinskis-theorem-regarding-darboux-functions – Martin Sleziak Aug 05 '14 at 08:41
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BTW isn't this a duplicate of this question: http://math.stackexchange.com/questions/179310/function-writen-as-two-functions-having-ivp? – Martin Sleziak Aug 06 '14 at 05:03
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As I have mentioned in a comment to an older question, this can be found as Theorem 9.5, p.57 in van Rooij, Schikhof: A Second Course on Real Functions. – Martin Sleziak Aug 06 '14 at 07:45
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Yes, it is true. Partition $I$ in two sets $A,I\setminus A$, both having cardinality $c$ and with the property that every interval in $I$ intersects both $A$ and $I\setminus A$. Let $g,h$ be two functions which take on every value in every relative subinterval of $A$ (and $I\setminus A$ respectively). We can extend $g$ to $I\setminus A$ and $h$ to $A$ in such a way that $g+h=f$ (by defining $g$ as $f-h$ on $I\setminus A$ and $h$ as $f-g$ on $A$) holds over the whole $I$. Since $g$ and $h$ obviously have the Darboux property, we are done.
Jack D'Aurizio
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Please elaborate the partitioning and how can we define and extend $g,h$ .... – Souvik Dey Aug 03 '14 at 04:08
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The extension is already given in the most explicit way (take $g$ as $f-h$ on $I\setminus A$ and $h$ as $f-g$ on $A$). The partitioning can be done by considering a Hamel basis (http://mathworld.wolfram.com/HamelBasis.html) and by putting into $A$ all (and only) the elements of $[0,1]$ such that $r_1=0$. – Jack D'Aurizio Aug 03 '14 at 04:28
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2You probably need not only that $A$ and $I\setminus A$ have cardinality $\mathfrak c$, but also that $A\cap(a,b)$ and the complement have the cardinality $\mathfrak c$ for every subinterval. – Martin Sleziak Aug 05 '14 at 08:44