let E be subset of R which has no isloated points(or C does not have any isolated point of E) and C be countable subset of R does there exist a monotonic function on E which is continuous only at points in E-C?

The problem is from Royden 4th edition page 109.

I know the proof in case E is an open bounded interval only.



$$L=\left\{x\in C:\exists y_x\in\Bbb R\big(y_x<x\text{ and }(y_x,x)\cap E=\varnothing\big)\right\}\;;$$

this is the set of points in $C$ that are not limits from the left of points in $E$. Another way to say it is that $x\in L$ if and only if $x\in C$ and $x>\sup_{\Bbb R}\{y\in E:y<x\}$. Let $Y=\{y_x:x\in L\}$; $C\cup Y$ is countable, so let

$$C\cup Y=\{x_n:n\in\Bbb N\}\;.$$

(I’m assuming that $C$ is countably infinite; if $C$ is finite, the problem is fairly trivial.) Let

$$f:E\cup Y\to\Bbb R:x\mapsto\sum_{x_n\le x}\frac1{2^n}\;.$$

Finally, let

$$g:E\to\Bbb R:x\mapsto\begin{cases} f(y_{x_n}),&\text{if }x=x_n\in Y\\ f(x),&\text{otherwise}\;. \end{cases}$$

The definition of $f$ ensures that $g$ is discontinuous from the left at every point of $C\setminus L$ and continuous everywhere else, and the modification to get $g$ ensures that $g$ is discontinuous from the right at every point of $L$ without affecting continuity at any other point of $E$. Thus, $g$ is discontinuous precisely at the points of $C$.

  • $\begingroup$ I again tried after discussing with you. Let min C doesn't exist. We arrange elements of C in sequence and define f as you have but taking C instead of C U Y. And in case min C exist, we arrange elements of C in sequence with x1 = minC. And defing g same as f except at minC we define it -1. I think it is working. Is there some problem in it? $\endgroup$ – Sushil Nov 26 '14 at 11:08
  • $\begingroup$ @Sushil: What if $E=[0,1]\cup[2,3]$, and $C=E\cap\Bbb Q$? Your function will almost work, but not quite, because it will be continuous at $2$. You need to decrease $g(2)$ to something bigger than $f(1)$ but less than $f(2)$. That’s what the points of $Y$ are for in my version. $\endgroup$ – Brian M. Scott Nov 26 '14 at 17:57
  • $\begingroup$ Thanks I didn't think about it. Okay regarding your proof now. I didn't get why g is continuous at x=xn∈Y $\endgroup$ – Sushil Nov 26 '14 at 18:24
  • $\begingroup$ And one more confusion in Y we are taking all y(x) with stated property not jsut one. Am I right? $\endgroup$ – Sushil Nov 26 '14 at 18:26
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    $\begingroup$ @Sushil: I don’t care whether I use AC, so I didn’t really think about it, but in fact you don’t need AC. If $x=\min E$, take $y(x)=x-1$. Otherwise, let $$y(x)=\frac12\big(x+\sup_{\Bbb R}\{z\in E:x<x\}\big)\;.$$ $\endgroup$ – Brian M. Scott Nov 26 '14 at 18:37

Yes, list $E\cap C = \{x_n:n=1,2..\}$. For each $x$ let $N_x=\{n:x_n<x\}$. Define $f(x)=\sum_{n\in N_x}2^{-n}$.

I do not quite see what is the role of $E$ in this question, we could define $f$ as above, initially disregarding $E$ (and using $C = \{x_n:n=1,2..\}$) and then later restricting this function to $E$.

Edit. This may not be "discontinuous enough" at points of $C$, as discussed in comments below. So define also $M_x=\{n:x_n\le x\}$, define $g(x)=\sum_{n\in M_x}2^{-n}$, and define $h(x)=f(x)+g(x)$. I hope $h$ works, if not then I do not understand the question and may need to read all over again.

Edit. $h$ is continuous at points of $E\setminus C$ (verify:). But $h$ (and any function) would be continuous at those points of $C$ that are isolated in $E$, that is at any $c\in C$ which has a neighborhood which misses all other points in $E$. It is ok if $c$ is isolated in $C$,but not in $E$, in that case $h$ would be discontinuous at $c$. Indeed $c=x_m$ for some $m$ (according to the above definition of $f$). So, $2^{-m}$ is not one of the members of the sum that defines $f(c)=f(x_m)=\sum_{n\in N_{x_m}}2^{-n}$ (this is since $m\not\in N_{x_m}$, since in the definition of $N_x$ we have strict inequality, but it is not true that $x_m<x_m$). On the other hand, if $x>x_m$, then $2^{-m}$ is one of the members of the sum that defines $f(x)=\sum_{n\in N_{x}}2^{-n}$, this is because $m\in N_x$, since $x_m<x$. It follows that if $x>x_m$ then $f(x)\ge f(x_m)+2^{-m}$, so $\lim_{x\to x_m^+} f(x)\ge f(x_m)+2^{-m}$, here we are assuming the there are points in $E$ to the right of $x_m$, arbitrarily close to $x_m$ (that is, we are assuming that $x_m$ is not isolated from the right in $E$). So $f$ has a jump at $x_m$ (to the right of $x_m$) so $f$ is discontinuous (from the right) at $x_m$. Similarly, if $x_m$ is not isolated from the left in $E$ then one may show that $\lim_{x\to x_m^-}f(x)\le f(x_m)-2^{-m}$, so $g$ has a jump at $x_m$ (at the left of $x_m$). So, if $x_m$ is not isolated in $E$ then it is either not isolated from the left, or not isolated from the right, or both, so either $g$ of $f$ or both are discontinuous at $x_m$, so $h$ is discontinuous at $x_m$. You should be able to verify all these details yourself, once you know what you are trying to prove.

  • $\begingroup$ What about E =[a, b] and when a is in C? $\endgroup$ – Sushil Nov 25 '14 at 1:10
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    $\begingroup$ The problem I see is at enpoints. will this function work on [0,1] with C as rational no. Why would it be discontinuous at 0? $\endgroup$ – Sushil Nov 25 '14 at 1:14
  • $\begingroup$ I guess you might want to use two jumps at each element of $C$, instead of just one as I did above. Use a jump at the left and a jump at the right, and as before these jumps may come from the set $2^{-n}$. $\endgroup$ – Mirko Nov 25 '14 at 1:15
  • $\begingroup$ No I don't really want to use two jumps. But function f defined by you would be discontinuous at x in C if C is in (0,1] this I can see. But would it be discontinuous at 0 with E = [0,1]. I just need explanation of this. I am not figure it out $\endgroup$ – Sushil Nov 25 '14 at 1:18
  • $\begingroup$ Sorry if I am not able to explain question properly. But you have got question right. I am not able to understand explanation. If I take h on [o,1] with C = Q in [0, 1]. Why would h be discontinuous at 0? $\endgroup$ – Sushil Nov 25 '14 at 1:27

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