# 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 $$f\colon \mathbb{R}\to\mathbb{R}$$ has the property that given any reals a < b, the function $$f\restriction [a,b]$$ 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 $$R\subset \mathbb{R}\times\mathbb{R}$$ be a relation such that $$R(x,y)\iff x-y\in \mathbb{Q}.$$ 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 $$x\in A$$ let us define the set $$A_x=\mathbb{Q} + x.$$. Clearly:

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

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

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

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

For details see this
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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