0
$\begingroup$

I was introduced in class to this function, described here in a post from a few years ago: https://math.stackexchange.com/a/44285/429158

This function has the intermediate value property, i.e. its image is an interval, but it is nowhere continuous.

I do get why it has the intermediate value property, but I can't wrap my head around why is it nowhere continuous. Any insights?

$\endgroup$
  • 1
    $\begingroup$ The very short answer is that it takes every value in $[0,1]$ within any interval with a nonempty interior. This simple point is perhaps easily overlooked in the mass of detail in that post. $\endgroup$ – Harald Hanche-Olsen May 31 '17 at 12:14
  • $\begingroup$ Sorry, I feel like a dummy but I still don't quite get it. Would you mind to elaborate a little? $\endgroup$ – Noa May 31 '17 at 12:20
2
$\begingroup$

To quote the question from the linked post:

Question: Recall that every $x \in (0,1)$ can be represented by a binary fraction $0.a_{1}a_{2}a_{3}\cdots$, where $a_{i} \in \{0,1\}$ and $i=1,2, \cdots$. Let $f: (0,1) \to [0,1]$ be defined by $$ f(x) = \overline{\lim_{n \to \infty}} \ \frac{1}{n} \sum\limits_{i=1}^{n}a_{i}$$ Prove that $f$ is discontinuous at each $x \in (0,1)$ but nevertheless has the intermediate value property.

Before continuing, I should mention a minor point: The function $f$ is not well defined, since any dyadic rational $x$ has two binary expansions, one ending in all zeros and one ending in all ones. So you get both $f(x)=0$ and $f(x)=1$ from the definition. This is easily repaired, though: Just choose one or the other. From a quick reading of the linked answer, I guess the choice is $f(x)=1$ for dyadic rationals $x$.

Now the point is this: To describe $x$ to great accuracy, you need to fix the first $k$ bits $a_i$, for some $k$. But once that is done, you can make $f(x)$ be whatever you want, just by adjusting the frequency of the following bits. It quickly follows that $f$ takes all values in $[0,1]$ in any open subinterval of $(0,1)$.

If $f$ is continuous at $x$, then there is some $\delta>0$ so that $|f(y)-f(x)|<\frac14$ whenever $|y-x|<\delta$. That is, $f(y)\in\bigl(f(x)-\frac14,f(x)+\frac14\bigr)$. But that interval has length $\frac12$, so it does not cover $[0,1]$. That is a contradiction, so $f$ is discontinuous.

The intermediate value property should be quite obvious.

For technical details, of course, consult the linked answer.

Edit: On second thought, I think the argument in the linked answer is more simply done as follows: Given $y\in[0,1)$, and assuming $a_i$ is selected for $i<n$, pick $a_n=1$ if that makes $n^{-1}\sum_{i=1}^n a_i<y$. Otherwise, pick $a_n=0$. Now the average $n^{-1}\sum_{i=1}^n a_i$ will grow with $n$ until it exceeds $y$, then it will shrink until it is less than $y$, and so forth, moving up and down in increasingly smaller excursions around $y$, and therefore converging to $y$. (Writing a formal proof is left to the reader.)

$\endgroup$
  • $\begingroup$ Incredible answer, I'm very grateful. $\endgroup$ – Noa Jun 1 '17 at 16:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.