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Let $C$ be a countable subset of $(a,b)$. Then there is an increasing function on $(a,b)$ that is continuous only on $(a,b)\setminus C$

This is an example from Royden's real analysis book. The function defined on $(a,b)$ is $f(x)=\sum_{\{n:q_n \le x\}}\frac{1}{2^n}$ where $\{q_n\}$ is an enumeration of $C$ . To show continuity on $(a,b)\setminus C$ he says let $x_0$ be in $(a,b)\setminus C$ and let $n$ be any natural number. Then there exists an interval, $I$, that contains $x_0$. In addition, $q_n$ is not in $I$ for $1\le k \le n$. Then this implys that $|f(x)-f(x_0)| < \frac{1}{2^n}$ for $x \in I$.

I understand this part. Then there is a problem right after this proof says:

  1. Let $C$ be a countable subset of the nondegenerate closed bounded interval $[a,b]$. Then there is an increasing function on $[a,b]$ that is continuous only on $[a,b]\setminus C$.

I was wondering, if we define $$f(a)=0 \text{ and } f(b)=\sum^{\infty}_{n=1}\frac{1}{2^n}=1$$ Is the proof still hold? Did I miss something here? Are the proof of these two problems similar?

Thank you!

Thank you for the solution I accepted, and I added something new here.

  1. Show that there is a strictly increasing function on $[0,1]$ that is continuous only at the irrationals in $[0,1]$.

  2. Let $f$ be a monotone function on a subset $E$ of $\mathbb R$. Show that $f$ is continuous except possibly at a countable number of points n $E$.

  3. Let $E$ be a subset of $\mathbb R$ and $C$ a countable subset of $E$. Is there a monotone function on $E$ that is continuous only at points $E \setminus C$?

These are four successive problems on Page 109 from Royden's real analysis book. So I think for problem 2, we can just use the result from problem 1 and let $C$ be the rationals. Did I miss something here?

And for 3 and 4, what are the difference between them? I mean if 3 is true which means we do have this function. Or did I misunderstand something here?

Thank you!

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  • $\begingroup$ My edit was to deleted the first "continuous" from the statement of the Q so that it makes sense. Apparently you made a "slip of the pen". $\endgroup$ – DanielWainfleet Nov 7 '17 at 19:23
  • $\begingroup$ @DanielWainfleet Thanks! $\endgroup$ – Nan Nov 7 '17 at 19:31
  • $\begingroup$ My second edit was the same as the first, but for the second proposition. $\endgroup$ – DanielWainfleet Nov 7 '17 at 19:54
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    $\begingroup$ I dk Royden's book or his usage of "countable"..... Some people (including me) prefer "countable" to mean "not uncountable" so that "countable" means "finite or countably infinite".... Of course, in this Q, the case of a finite $C$ is obvious. $\endgroup$ – DanielWainfleet Nov 7 '17 at 20:19
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For the case $[a,b],$ the function $f$ is continuous at $a,$ but it may be that $a\in C.$

If $a\not\in C $ let $g=f$.

If $a\in C $ let $g(a)=-1$ and $g(x)=f(x)$ for $x\in (a,b]. $

Then $g$ is continuous only on $[a,b]$ \ $C.$

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HINT: $f$ must not be continuous at $\{a,b\}\cap C$.

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  • $\begingroup$ Thank you for replying. Sorry I did not quite get your hint. Since $f$ is not continuous on $C$, and $\{a,b\}\cap C \subseteq C$, so $f$ is not continuous on $\{a,b\}\cap C$ for sure. $\endgroup$ – Nan Nov 7 '17 at 18:53
  • $\begingroup$ The hint was meant to say that you have to make sure that $f$ is not continuous at $a$, nor at $b$, which the "basic" construction you gave does not ensure. Note that, as far as I can tell, the "accepted" answer still fails to work for $b\in C$ if the set of discontinuity points converges to $b$. But it's easy to take a function and add discontinuities at a finite number of points without "substantially" changing it! $\endgroup$ – Anonymous Nov 8 '17 at 22:06

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