Let $\{F_n\}_{n\geq1}$ be a sequence of closed sets such that $F_n\subseteq F_{n+1}$ for all $n\geq1$. Let $F=\cup_{n\geq 1}F_n$ and $F_0=\emptyset$. For $n\geq 1$ we define $A_n=[(F_n\setminus F_{n-1})\setminus Int(F_n\setminus F_{n-1})]\cup [Int(F_n\setminus F_{n-1})\cap Q]$. Let $f:R\to R$ given by $f(x)=2^{-n}$ if $x\in A_n$, $f(x)=0$ if $x\notin \cup _{n\geq 1}A_n$. Show that $f$ is discontinuous on $F$ and continuous on $R\setminus F$.

I don't even know how to start with this thing. I suppose continuity is easier to show? So I pick $x\in R\setminus F$, I need to find $\delta>0$ such that $f(B_{\delta}(x))\subseteq B_{\epsilon}(0)$. Considering infinite union of closed sets can be non-closed, I would have to enter the territory of $F$. Since $F_n$ is nested, we can go in a little bit of $F$ and the value there should still be small? How do I write this in math? And I don't have much clue about discontinuity.


For the proof that $f$ is continuous on $\mathbb{R} \backslash F$ one can formalize your (good) intuition as follows: fix $\varepsilon > 0$ and $x_0 \in \mathbb{R} \backslash F$. We must show that there is $\delta >0$ so that if $|x - x_0| \leq \delta$, then $|f(x) - f(x_0)| \leq \varepsilon$. We note that $f(x_0) = 0$ because $\mathbb{R} \backslash F \subset \mathbb{R} \backslash A$. Thus, we must find an interval around $x_0$ on which $f$ is small. For this, let $N$ be such that $2^{-N} < \varepsilon$. Since each $F_n$ is closed, the finite union $\cup_{n \leq N} F_n$ is closed, which means its complement is open. Consequently, we can choose $\delta > 0$ so that
$$(x_0, - \delta, x_0 + \delta) \subset \mathbb{R} \backslash (\cup_{n \leq N} F_n) \subset \mathbb{R} \backslash (\cup_{n \leq N} A_n),$$ which means that for $x \in (x_0, - \delta, x_0 + \delta)$, we have $$ |f(x) - f(x_0)| = |f(x)| < 2^{-N} \leq \varepsilon. $$

To prove that $f$ is discontinuous on $F$, we will show that if $x_0 \in \mathbb{R}\backslash F$, then there exists $\varepsilon_0 > 0$ so that for any $\delta > 0$, there is $x \in \mathbb{R}$ with $|f(x) - f(x_0)| \geq \varepsilon_0$. If $F$ is empty, we are done. Otherwise, let $x_0 \in F$ be fixed. There must be some $n \geq 1$ so that $x_0$ is in $F_n$ but not in $F_{n-1}$. For such an $x_0$, we can take $\varepsilon_0 = 2^{-n}$. This is because:

(a) if $x \in A_n$, then any open interval around $x_0$ contains a member of $F_{n} \backslash F_{n-1}$ that is irrational, and

(b) if $x_0 \in F_{n} \backslash F_{n-1} $ but not in $A_n$, then any open interval around $x_0$ contains a member of $A_n$.
(I can provide more details for these last two assertions if you wish.)

  • $\begingroup$ (a) is due to irrationals is dense in R. (b) is saying An closure contains Fn\Fn-1. (b) should be caused by rationals is dense in R but I have trouble writing (b) down. Could you please explain more about (b)? $\endgroup$ – Fluffy Skye Feb 6 '19 at 18:21
  • $\begingroup$ Actually, (a) is due to density of irrationals and (b) is due to density of rationals. If $x_0 \in F_n \backslash F_{n-1}$ but not in $A_n$, then $x_0$ is an irrational number in the interior of $F_n \backslash F_{n-1}$. Any open interval around $x_0$ contains a rational number that is also in the interior of $F_n \backslash F_{n-1}$ and hence a member of $A_n$. $\endgroup$ – Jordan Green Feb 6 '19 at 18:26
  • $\begingroup$ Ok. So there are two cases for (a). If x is in interior of Fn\Fn-1. Then x is rational. Then any small open ball around it contains irrationals who are not in An and we're good. But what if it's not in interior of Fn\Fn-1? $\endgroup$ – Fluffy Skye Feb 6 '19 at 22:46
  • 1
    $\begingroup$ @FluffySkye If $x \notin \operatorname{int} F_n \setminus F_{n-1}$, then $B_r(x)$ is not contained in $F_n \setminus F_{n-1}$ for any $r > 0$. Given $f(x) = 2^{-n}$, we can consider $\varepsilon = 2^{-n-1} > 0$, and, assuming continuity at $x$, find $\delta > 0$ such that $$|y - x| < \delta \implies |f(y) - f(x)| < \varepsilon.$$ If you work this out, you'll find that $2^{-n-1} < f(y) < 2^{-n+1}$, which means $f(y) = 2^{-n}$. This can only happen if $y \in F_n \setminus F_{n-1}$, hence $B_\delta(x) \subseteq F_n \setminus F_{n-1}$. This is a contradiction. $\endgroup$ – Theo Bendit Feb 7 '19 at 2:04

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.