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Let $r_1, r_2,.... $ be an enumeration of the set of rationals on the real line. For each $x\in \mathbb{R}$, let us put $f(x) = \sum_{n;r_n\leq x} \dfrac{1}{2^n} $ . Show $f$ is continuous at a point $a\in [0,1]$, iff $a$ is irrational.

for irrational points the terms of the series are infinite and for rational points the terms are finite and the sums are finite. So how do I choose an $\epsilon$ to prove that $f$ is not continuous at $a\in \mathbb{Q}$? Any hints?

Amirhossein
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Lucas
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  • I don't know then, I found it in a problem set. – Lucas Oct 02 '20 at 20:18
  • $(0,1)$ is the domain and I added a full stop there. I think it make sense now – Lucas Oct 02 '20 at 20:19
  • @uniquesolution I intepret that statement as: If $a \in (0,1)$ then $f$ is continuous at $a$ if and only if $a$ is irrational". I'd say that what is written is ungrammatical and contains a typo but it is salvageable. (If my interpretation is correct,, that is, ... I don't want to make any unwarranted assumptions.) – fleablood Oct 02 '20 at 20:27
  • "for irrational points the terms of a series are infinite and for rational points the terms are finite and the sums are finite" I don't think that is true for any $x \in (0,1)$ whether $x$ is rational or irrational there are any infinite number or $r_n < x$ and and infinite number of $r_m > x$. The only difference is if $x$ is rational then there is an $r_n = x$ but there isn't if $x$ is irrational. Hint: (I think) If $x$ is rational then for every $\epsilon$, then $r_n=x$ and $n$ in the index. If $x$ is irrational the you can force all the indexes to be high. – fleablood Oct 02 '20 at 20:39
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    "for rational x once it reaches x the index stops. Right?" No. if $r_m = q$ then you don't stop. There will be $n > m$ so that $r_n < r_m = q$. As there are infinitely many rationals less than $q$ there are infinitely many $r_n < q$ and infinitely many of the those $n$ will be greater than $m$. – fleablood Oct 02 '20 at 20:48
  • @fleablood Oh!! yeah, I'll try – Lucas Oct 02 '20 at 20:50
  • @fleablood can you please elaborate on "If x is rational then for every ϵ, then $r_n=x$ and n in the index." this n won't affect the sum, I can't find any contradiction – Lucas Oct 02 '20 at 21:01
  • Try showing that if $x$ is rational, $x = r_k$, then $$f(x) - \lim_{\substack{ y \to x \ y < x}} f(y) = \frac{1}{2^k},.$$ – Daniel Fischer Oct 03 '20 at 10:19
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    Claim: If you have a point $x$, and a very large natural natural number $N$, you can find an $\epsilon > 0$ so that for all the rationals $q=r_n$ (other than $x$) so that $0 < |x-q|< \epsilon$ that the index of $q = r_n$, that is $n$, will be that $n > N$.... Pf: Let $\epsilon = \min(|x - r_k|: k \le N, r_k \ne x)$...... Can you use that? – fleablood Oct 03 '20 at 16:16
  • See also: Construct a monotone function which has countably many discontinuities and Prove $f$ is discontinuous on $r_n$? – Martin Sleziak Oct 08 '20 at 17:24

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Note that if $x < y$ then if $D_x= \{m| r_m \le x\}$ and $D_y = \{m|r_m\le y\}$ then $D_x \subset D_y$. And as there is rational $q$ so that $x < q < y$ and there is a $k$ so that $q = r_k$ so $k\in D_y$ but $k\not \in D_x$, $D_x$ is a proper subset of $D_y$. So $f(x) =\sum_{m\in D_x}\frac 1{2^m} <\sum_{m\in D_y}\frac 1{2^m} = f(y)$.

Now $q= r_k$ is rational then for all $x < r_k$ then $k \not \in D_x\subsetneq D_q$ but $k \in D_q$ so $f(x) = \sum_{m\in D_x} f(x) \le \sum_{m\in D_q;m\ne k} \frac 1{2^m} < \sum_{m\in D_q;m\ne k} \frac 1{2^m}+\frac 1{2^k} =\sum_{m\in D_q}\frac 1{2^m} =f(q)$. So $|f(q) - f(x)| \ge \frac 1{2^k}$. So $f$ can not be continuous at any rational $q$.

Now suppose $x$ is irrational and let $D_x = \{m| r_m \le x\}$.

Now for any $\epsilon > 0$ then let $\frac 1{2^w} < \epsilon$. $\frac 1{2^w} = \sum_{j> w} \frac 1{2^j}$.

Now let $\delta = \min_{m=1}^w(|x-r_m|)$.

Now the only rationals in the interval $(x - \delta, x + \delta)$ can only have indexes larger that $w$ (because if $q = r_k$ and $k \le w$ then $|x-r_k| \ge \delta$ by the way we defined $\delta$).

Now if $|x-y| < \delta$ then only the elements in $D_y$ that are not in $D_x$ (if $y > x$) or the elements of $D_x$ not in $D_y$ (if $x > y$) are the indexes of rationals in the interval so the only elments of $D_y$ not in $D_x$ (or vice versa) are indexes larger than $w$.

So $|f(x) - f(y)| \le \sum_{j>w} \frac 1{2^j} = \frac 1{2^w} < \epsilon$.

So $f$ is continuous at $x$.

fleablood
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