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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.

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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.)

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  • $\begingroup$ (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
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    $\begingroup$ @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

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