Solution To The Previous Problem

Exceptionally I will present a solution to the previous Wednesday Problem. Only not shall I present any simplest solution but a cool one.

The question was the following.

Suppose a function [tex]f\colon \mathbb{R}\to\mathbb{R}[/tex] has the property that given any reals a < b, the function [tex]f\restriction [a,b][/tex] takes all possible values between f(a) and f(b). Call this property iv-property (for intermetiate value). Can a function with iv-property be discontinuous?

The solution I give here shows that it can be discontinuous at every point.

(And not measurable).

Let [tex]R\subset \mathbb{R}\times\mathbb{R}[/tex] be a relation such that [tex]R(x,y)\iff x-y\in \mathbb{Q}.[/tex] This is clearly an equivalence relation. Using Axiom of Choice we can make up a set A whose intersection with each equivalence class of R is a singleton. Such A is known as the non-measurable Vitali set.

For each [tex]x\in A[/tex] let us define the set [tex]A_x=\mathbb{Q} + x.[/tex]. Clearly:

(1) If [tex]x\ne y[/tex] are in A, then [tex]A_x\cap A_y=\varempty.[/tex]
(2) [tex]A_x[/tex] is countable and dense subset of [tex]\mathbb{R}[/tex].
(3) [tex]\mathbb{R}=\bigcup_{x\in A}A_x.[/tex]
(4) Cardinality of A is the same as that of real numbers: [tex]|A|=|\mathbb{R}|.[/tex]

By (4) there is a bijection [tex]g\colon A\to \mathbb{R}:[/tex] with each set [tex]A_x[/tex] we associate the a unique real number [tex]g(x)[/tex]. Let us define the function [tex]f\colon \mathbb{R}\to\mathbb{R}[/tex] as follows: if [tex]y\in A_x[/tex], then let [tex]f(y)=g(x).[/tex] This is well defined map: by (3), y is in some [tex]A_x[/tex] and by (1) there is only one possible choice for [tex]f(y)[/tex].

Let us show next that f has the iv-property. Assume that a<b are real numbers. If [tex]f(a)=f(b)[/tex], then there is nothing to prove. Suppose that [tex]f(a)<f(b)[/tex] and let [tex]x\in [f(a),f(b)].[/tex] Let [tex]z = g^{-1}(x).[/tex] By (2) the set [tex]A_z[/tex] is dense in [tex]\mathbb{R}[/tex], so there must be a [tex]y\in A_z\cap [f(x),f(y)].[/tex] By definition [tex]f(y)=x.[/tex]

Let us now show that if [tex]x_0\in \mathbb{R}[/tex] is arbitrary, then [tex]f[/tex] is not continuous at [tex]x_0.[/tex] This is easy, since [tex]A_y=A_{x_0+\sqrt{2}}[/tex] is dense in any neighbourhood of [tex]x_0[/tex] and [tex]f(x_0)\ne f(y)[/tex] plus [tex]f[/tex] is constant on [tex]A_y[/tex].

About Vadim Kulikov

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3 Responses to Solution To The Previous Problem

  1. Anonymous says:

    I think it is just to mention the fact, that functions with iv-property are called (at least in some languages) Darboux functions and this is due to French mathematician Jean Gaston Darboux (btw doctoral advisor of Borel).
    http://en.wikipedia.org/wiki/Jean_Gaston_Darboux

  2. vadik says:

    Thank you, Anonymous. In our department the property would (and was) be called Bolzano property after mathematician Bernard Bolzano, because we call Intermediate Value Theorem Bolzano Theorem. And I knew that in some countries this theorem is called Darboux theorem, though I did not remember this name. So thank you for reminding.

    According to Wikipedia, however, Bolzano theorem is the intermediate value theorem for continuous functions and Darboux theorem is the intermediate value theorem for functions that result from differentating differentiable functions.

    http://en.wikipedia.org/wiki/Darboux%27s_theorem_(analysis)
    http://en.wikipedia.org/wiki/Bolzano%27s_theorem

    I guess it could be the Bolzano-Darboux property after all? Also iv-property seems loyal to me :)

  3. Pingback: Walks On Math » Around Jordans Curve Theorem I

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