Analysis? Set theory?

For a subset of real numbers \(S\), let \(\mathbf{1}_S : \mathbb{R} \to \{0, 1\}\) be the indicator function of SS, defined as 1S(x)=1\mathbf{1}_S(x) = 1 if xSx \in S and 1S(x)=0\mathbf{1}_S(x) = 0 otherwise.

Prove or disprove: for every real function f:RRf : \mathbb{R} \to \mathbb{R}, there exists subsets of real numbers A1,A2,A3,A_1, A_2, A_3, \ldots and real numbers c1,c2,c3,c_1, c_2, c_3, \ldots such that

f(x)=n=1cn1An(x)\displaystyle\large{ f(x) = \sum_{n=1}^\infty c_n \mathbf{1}_{A_n}(x) }

for all real xx.


Clarification: When I posted this problem, I didn't know the answer. Now, I found the answer, but I find it interesting (like most set theory stuff), so I'll let you to figure it out.

#Algebra

Note by Ivan Koswara
5 years, 1 month ago

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Comments

Given a real number, we can express its binary representation as b2b1b0.b1b2\dots b_2 b_1 b_0.b_{-1} b_{-2} \dots, where each digit bib_i is 0 or 1.

For an integer ii, let SiS_i be the set of real numbers xx such that the iith digit in the binary representation of f(x)f(x) is equal to 1. Then f(x)=iZ2i1Si(x).f(x) = \sum_{i \in \mathbb{Z}} 2^i \mathbf{1}_{S_i}(x).

Jon Haussmann - 4 years, 7 months ago
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