- #1

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- Thread starter Dragonfall
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- #1

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- #2

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Sure there is.

- #3

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Can you give an example?

- #4

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Well....the irrationals aren't countable...

- #5

matt grime

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1) continuous

2) non-constant

3) 0 on [0,1]

- #6

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He said that f is nowhere constant.

- #7

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Yes, I did say f is **nowhere constant**. If f=0 on the subset of irrationals of some interval, then continuity implies that f=0 on that interval.

Intuitively, what I want is a everywhere continuous nowhere differentiable function that is "straight" enough so that a horizontal line intersects the values uncountably many times. I don't think such a function exists, but I can't prove it either way.

The "continuous but nowhere differentiable" requirement might not be necessary, or even relevant, but it's a good place to start looking.

Intuitively, what I want is a everywhere continuous nowhere differentiable function that is "straight" enough so that a horizontal line intersects the values uncountably many times. I don't think such a function exists, but I can't prove it either way.

The "continuous but nowhere differentiable" requirement might not be necessary, or even relevant, but it's a good place to start looking.

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- #8

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- #9

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Whoops. Missed the continuity assumption.

- #10

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Can you please "formally" define your function? "if we infinitely repeat..." is not a formal term and I am not sure the ultimate function remains a continuous one.

- #11

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"if we infinitely repeat..." simply refers to the limit function of the sequence of functions I described.

Define

[tex]

f_1[a,b](x) = \left\{

\begin{array}{cc}

0 & x \in \left[a,\frac{2a+b}{3}\right]\cup\left[\frac{a+2b}{3},b\right]\\

x-\frac{2a+b}{3} & x \in \left[\frac{2a+b}{3},\frac{a+b}{2}\right]\\

\frac{a+2b}{3}-x & x \in \left[\frac{a+b}{2},\frac{a+2b}{3}\right]

\end{array}

[/tex]

and

[tex]

f_{n+1}[a,b](x) = \left\{

\begin{array}{cc}

f_n\left[a,\frac{2a+b}{3}\right](x) & x \in \left[a,\frac{2a+b}{3}\right]\\

x-\frac{2a+b}{3} & x \in \left[\frac{2a+b}{3},\frac{a+b}{2}\right]\\

\frac{a+2b}{3}-x & x \in \left[\frac{a+b}{2},\frac{a+2b}{3}\right]\\

f_n\left[\frac{a+2b}{3},b\right](x) & x \in \left[\frac{a+2b}{3},b\right]

\end{array}

[/tex]

Then [tex]f(x) = \lim_{n\to\infty}f_n[0,1](x)[/tex] is the function I just described.

Define

[tex]

f_1[a,b](x) = \left\{

\begin{array}{cc}

0 & x \in \left[a,\frac{2a+b}{3}\right]\cup\left[\frac{a+2b}{3},b\right]\\

x-\frac{2a+b}{3} & x \in \left[\frac{2a+b}{3},\frac{a+b}{2}\right]\\

\frac{a+2b}{3}-x & x \in \left[\frac{a+b}{2},\frac{a+2b}{3}\right]

\end{array}

[/tex]

and

[tex]

f_{n+1}[a,b](x) = \left\{

\begin{array}{cc}

f_n\left[a,\frac{2a+b}{3}\right](x) & x \in \left[a,\frac{2a+b}{3}\right]\\

x-\frac{2a+b}{3} & x \in \left[\frac{2a+b}{3},\frac{a+b}{2}\right]\\

\frac{a+2b}{3}-x & x \in \left[\frac{a+b}{2},\frac{a+2b}{3}\right]\\

f_n\left[\frac{a+2b}{3},b\right](x) & x \in \left[\frac{a+2b}{3},b\right]

\end{array}

[/tex]

Then [tex]f(x) = \lim_{n\to\infty}f_n[0,1](x)[/tex] is the function I just described.

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- #12

Hurkyl

Staff Emeritus

Science Advisor

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There's a simpler description of (something like) Moo's function:

f(x) = [distance from x to the Cantor set]

Incidentally, I don't think I've heard "nowhere constant" before -- I would have used the phrase "locally nonconstant".

f(x) = [distance from x to the Cantor set]

Incidentally, I don't think I've heard "nowhere constant" before -- I would have used the phrase "locally nonconstant".

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- #13

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"if we infinitely repeat..." simply refers to the limit function of the sequence of functions I described.

Define

[tex]

f_1[a,b](x) = \left\{

\begin{array}{cc}

0 & x \in \left[a,\frac{2a+b}{3}\right]\cup\left[\frac{a+2b}{3},b\right]\\

x-\frac{2a+b}{3} & x \in \left[\frac{2a+b}{3},\frac{a+b}{2}\right]\\

\frac{a+2b}{3}-x & x \in \left[\frac{a+b}{2},\frac{a+2b}{3}\right]

\end{array}

[/tex]

and

[tex]

f_{n+1}[a,b](x) = \left\{

\begin{array}{cc}

f_n\left[a,\frac{2a+b}{3}\right](x) & x \in \left[a,\frac{2a+b}{3}\right]\\

x-\frac{2a+b}{3} & x \in \left[\frac{2a+b}{3},\frac{a+b}{2}\right]\\

\frac{a+2b}{3}-x & x \in \left[\frac{a+b}{2},\frac{a+2b}{3}\right]\\

f_n\left[\frac{a+2b}{3},b\right](x) & x \in \left[\frac{a+2b}{3},b\right]

\end{array}

[/tex]

Then [tex]f(x) = \lim_{n\to\infty}f_n[0,1](x)[/tex] is the function I just described.

Hmm.. each f_n is uniformly continuous, then if the seq {f_n} converges uniformly, the trick is done.

For the triangular waves become more little as n surges up, I think the convergence is uniform. But, can you dfevise a formal, maybe inductive, proof?

- #14

AKG

Science Advisor

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If C is a closed uncountable set in **R** containing no intervals (like the Cantor set), then it's complement is a countable union of disjoint open intervals. Define f to be 0 on C, and f is a triangle of slope with absolute value 1 on each open interval. So let C be the Cantor set. [itex]\mathbb{R} = (-\infty ,0) \sqcup \bigsqcup (a_n,b_n) \sqcup C \sqcup (1,\infty )[/itex]. Define [itex]f : \mathbb{R} \to \mathbb{R}[/itex] by:

[tex]f(x)=\left\{\begin{array}{cc}x,&\mbox{ if }

x < 0\\ \frac{b_n-a_n}{2} - |x - \frac{b_n+a_n}{2}|, & \mbox{ if } a_n < x < b_n\\ 0, & \mbox{ if } x \in C\\ 1-x, & \mbox{ if } x > 1\right.[/tex]

[tex]f(x)=\left\{\begin{array}{cc}x,&\mbox{ if }

x < 0\\ \frac{b_n-a_n}{2} - |x - \frac{b_n+a_n}{2}|, & \mbox{ if } a_n < x < b_n\\ 0, & \mbox{ if } x \in C\\ 1-x, & \mbox{ if } x > 1\right.[/tex]

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