Continuity of function

Suppose $f(x) = \begin{cases} 0 \ \ \text{if} \ x \in \mathbb{R}- \mathbb{Q} \newline \frac{1}{q} \ \ \text{if} \ x \in \mathbb{Q} \ \text{and} \ x = \frac{p}{q} \ \text{in lowest terms} \end{cases}$

(i) Is $f$ continual on the irrationals? (ii) Is $f$ continual on the rationals?

For (i) you could make use of the series definition of connection? Possibly attempt $a_n = \frac{\sqrt{2}}{n}$ and also show that $a_n \to 0$ yet $f(a_n) \not \to 0$? So its alternate on the irrationals?

For (ii) I do not see why there are $1+2+ \cdots + (q-1)$ sensible numbers? I recognize that we require to utilize this reality to pick an ideal $\delta$ (e.g $0< |x-a| < \delta \Rightarrow |f(x)-L| < \epsilon$).

2019-05-18 21:46:07
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Answers: 3

Your $a_n$ is are illogical, so $f(a_n)=0$. The function is continual on the irrationals. You can show this making use of the $\epsilon$ - $\delta$ definition of connection. As soon as you are offered an illogical $x$ and also an $\epsilon>0$, there is an integer $n>1/\epsilon$, and also there are just finitely several sensible numbers in, claim, $(x-1,x+1)$ having smaller sized than $n$ is cheapest terms. Hence there is a closest one to $x$, and also you can utilize this to locate your $\delta$.

Making use of comparable thinking, you can show that the restriction at each sensible is 0, yet, as ADVERTISEMENT. mention, without recognizing this there is a fast means to see that the function is alternate at each sensible number.

2019-05-21 07:50:16

Here is an image that could aid you see points even more plainly:

2019-05-21 07:49:31

It is an well well-known function, continual in all irrationals and also alternate in all rationals. I will certainly offer illustration of the evidence listed below.

First allow is see why it is continual at irrationals. To be continual at an illogical factor $x_0$, we require to show that $\forall\delta>0\exists\epsilon>0:|x-x_0|<\epsilon\Rightarrow |f(x)-f(x_0)|<\delta$.

Notification that within any kind of series of size 1, specifically within $[x_0-1/2,x_0+1/2)$, $f(x)$ takes the value 1/ 2 specifically as soon as, 1/ 3 two times, 1/ 4 thrice, and more, taking 1/k specifically k+1 times. So if we remove these factors upto claim $\lceil1/\delta\rceil$, we will certainly be getting rid of just finitely several factors, and also we will certainly be entrusted factors on which $f(x)<1/\lceil1/\delta\rceil\le\delta$. Notification that $x_0$ being illogical was not one of the factors that was gotten rid of. So locate the closest factor $a$ to $x_0$ that was gotten rid of, and also call the range $|x_0-a|$ as $\epsilon$.

Similarly, one can say that the function is alternate for rationals. If $x_1$ is a sensible factor, notification that if we remove all the factors of the kind $p/q$ upto $0\lt q\le N$, any kind of period around $x_1$ will certainly have all factors with $f(x)\lt 1/N$. So we can create periods around $x_1$ with randomly tiny $f$. Therefore $f$ is alternate at $x_1$ given that $f(x_1)>0$.

2019-05-21 07:43:21