0
$\begingroup$

Let $$f(x) = \cases{1/2^n & if $x = b/2^n$ with $b$ odd \cr 0 & otherwise}$$

Prove that $f$ is discontinuous at the points $x=b/2^n$ with $b$ odd, but is continuous at every other point.

Let's see, if I take $S$ to be the set of all numbers of the form $b/2^n$ then for $a \in S$ I'll let $0< \varepsilon < 1/2^n$ and $ \delta>0$ such that $\exists x\in \mathbb{R} / S $ : $|a-x|<\delta$. Then $|f(a)-f(x)|=1/ 2^n> \varepsilon$ so $f$ is discontinuous at $a$. Is this right? Now how to prove continuity at every other point?

$\endgroup$
  • $\begingroup$ You should distinguish two cases: when $x=0$ and when $x \ne 0$ (and not of the form $b/2^n$ for odd $b$).. For the latter case try to find an interval around each such $x$ so that in that interval $f(x)=0$. For the $x=0$ case, try to find the maxima of the $f(x)$ function within any interval around $0$. $\endgroup$ – Zoltan Zimboras Oct 22 '17 at 4:36
1
$\begingroup$

The numbers in the form of $b/2^n$ is called dyadic. Let $x$ be any point not dyadic. Given a certain $\epsilon$, we can choose $\delta$ small enough that in the interval $(x-\delta, x+\delta)$, the dyadic numbers all do not have small denominators.

Formally, let $N$ be a number big enough such that $2^N > 1/\epsilon$, and let $\delta_1 = 1/2^{N+1}$.

And let $\delta_2$ be a number small enough that the interval $(x-\delta_2, x+\delta_2)$ do not contain a dyadic number in the form of $b / 2^k$ where $k \leq N$.

For example, the interval $(3/8, 5/8)$ contains $1/2$ which has denominator $2^1$. Or another example, $(3/16,5/16)$ contains $1/4$ which has denominator $2^2$. If our choice of $\delta_1$ results in one of these cases, then we simply choose $\delta_2$ to be small enough to "avoid" stepping on the.se low-denominator dyadic number. To prove that this choice of $\delta_2$ is always possible, look at the binary expansion of $x$. Because $x$ is not dyadic, the binary expansion of $x$ will not terminate, but rather it goes on forever (it might repeat itself after a while but it will go on forever). Suppose the dyadic number we "stepped on" has form $b/2^k$ then just go further than the $k$-th digit.

Now it's rather straightforward to show that $f(x)$ in the interval $(x-\delta_2, x+\delta_2)$ will have maximum value $1/2^{N+1} < \epsilon$.

$\endgroup$
0
$\begingroup$

At every other point (when $x \in \mathbb{r} - S $), we have f(x)=0.

Let $\epsilon > 0$,
let $\delta$ be epsilon. $$ \forall x: 0<\lvert x-a \rvert < \delta,$$ $$ f(x)=0<\epsilon$$ $\square$ Q.E.D.

$\endgroup$

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.