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].