# Popcorn function

I need to construct a function whose set of points of discontinuities is a given F-sigma set.

At the Follow-Up section of Wikipedia's article on the popcorn function, they give an example of such a function. I just don't see why would we need closedness of the sets in the proof. Could somebody briefly explain the proof?

Thanks

• Please don't use abbreviations like "smb". The effort of a handful of keystrokes you save stands in no relationship to the effort you cause for everyone who tries to understand your post and isn't aware of the abbreviation. Sep 28, 2011 at 21:53
• – t.b.
Sep 28, 2011 at 22:16
• Pete's answer in the linked thread should explain that fully; keep in mind that the complement of an $F_\sigma$-set is a $G_{\delta}$-set.
– t.b.
Sep 28, 2011 at 22:17

For the record, $A=\bigcup_{n=1}^\infty F_n\subseteq\mathbb{R}$ where each $F_n$ is closed and $$f_A(x)= \begin{cases} 1/n, & \text{if } x \text{ is rational and } n \text{ is minimal so that }x\in F_n \\ -1/n, & \text{if } x \text{ is irrational and } n \text{ is minimal so that }x\in F_n \\ 0, & \text{if } x\notin A. \end{cases}$$ The closedness of $F_n$'s is needed for $f_A$ to be continuous in $\mathbb{R}\setminus A$. Let's look more closely at this.

Continuity at each point of $\mathbb{R}\setminus A$: Pick $x\in\mathbb{R}\setminus A$. Then $f(x)=0$. Let $\epsilon>0$ and pick $n\in\{1,2,\ldots\}$ so that $1/n<\epsilon$. Because each of the sets $F_1,\ldots,F_n$ are closed, we can choose $\delta>0$ so that $(x-\delta,x+\delta)$ does not intersect any of the sets $F_1,\ldots,F_n$. But then $$|f(y)-f(x)|=|f(y)|<1/n<\epsilon\quad\text{for all } y\in(x-\delta,x+\delta).$$ Hence $f_A$ is continuous at $x$.

For completeness:

Discontinuity at each point of $A$: Pick $x\in A$. For simplicity assume that $x$ is rational (the other case is very similar). Then $f(x)=1/n$ for some $n\in\{1,2,\ldots\}$. Let $\delta>0$. Now there is an irrational number $y$ in $(x-\delta,x+\delta)$. If $y\in\mathbb{R}\setminus A$, then $f(y)=0$; if $y\in A$, then $f(y)<0$. In any case $f(y)\leq 0$, hence $|f(y)-f(x)|\geq 1/n$. This means $f_A$ is discontinuous at $x$.

It’s convenient to assume that the closed sets $F_n$ are increasing, so that $F_1 \subseteq F_2 \subseteq F_3 \subseteq \dots$. This is a harmless assumption, since the union of any finite number of closed sets is closed: just replace $F_n$ by $\bigcup_{i=1}^n F_i$. Now we have $A = \bigcup_{n=1}^\infty F_n$, and we define the function $$f_A(x) = \begin{cases} \frac1n,&\text{ if }x\text{ is rational and }n\text{ is minimal so that }x\in F_n\\ \frac{-1}{n}&\text{ if }x\text{ is irrational and }n\text{ is minimal so that }x\in F_n\\ 0,&\text{ if }x \notin A. \end{cases}$$

First we show that $f_A$ is continuous at each point of $\mathbb{R}\setminus A$. Suppose that $x \in \mathbb{R}\setminus A$; clearly $f_A(x)=0$. If $x$ has a nbhd $V$ disjoint from $A$, then $f_A(y)=0$ for every $y \in V$, so $f_A$ is certainly continuous at $x$. Assume now that $x$ has no such nbhd, so that every nbhd of $x$ meets $A$.

For $n \in \mathbb{Z}^+$ let $V_n = \mathbb{R}\setminus F_n$; each $V_n$ is a nbhd of $x$. Suppose that $y \in V_m$ for some $m$. If $y \notin A$, then $f_A(y) = 0$. Otherwise, $f_A(y) = \pm 1/n$, where $n$ is minimal with $y \in F_n$. Since $y \in V_m$, $y \notin F_m$; and since the $F_i$ are nested, $y \notin F_i$ for any $i \le m$. Thus, $n>m$. This shows that $$\vert f_A(y)\vert < \frac1m$$ for every $y \in V_m$. In other words, given any $\epsilon > 0$, we can choose a positive integer $m$ such that $1/m < \epsilon$, and $V_m$ will be a nbhd of $x$ such that $\vert f_A(y)-f_A(x)\vert < \epsilon$ for every $y \in V_m$. This of course means that $f_A$ is continuous at $x$.

It remains to show that $f_A$ is discontinuous at each point of $A$. Fix $x\in A$, and assume that $x$ is rational. (The argument for irrational $x$ is almost identical.) Then $f_A(x) = 1/n$ for some $n \in \mathbb{Z}^+$, and we have to consider two possibilities. First, it may happen that $x$ has a nbhd $V\subseteq F_n$. If $n>1$, we may further assume that $V \subseteq V_{n-1}$, since $x$ is not in the closed set $F_{n-1}$. (Here I’m using the fact that $F_{n-1}$ is closed.) Let $W$ be any nbhd of $x$; then $W\cap V$ is a nbhd of $x$, so it contains some irrational $y$. But $W \cap V \subseteq F_n$, so $y\in F_n$; moreover, $y\in V_{n-1}$ if $n>1$, so $f_A(y) = -1/n$. Thus, each nbhd of $x$ contains a point $y$ such that $$\vert f_A(y)-f_A(x)\vert = \left\vert\frac1n-\frac{-1}n\right\vert=\frac2n,$$ and $f_A$ must be discontinuous at $x$.

The other possibility is that no nbhd of $x$ is contained in $F_n$. Then if $V$ is a nbhd of $x$, $V\cap V_n \ne \varnothing$. Let $y \in V\cap V_n$ be irrational. (Here again I’m using the fact that $F_n$ is closed: I need to know that $V\cap V_n$ is open in order to be sure that it contains an irrational.) If $y\in A$, $f_A(y)<0$, and if $y\in \mathbb{R}\setminus A$, $f_A(y)=0$, so in any case $f_A(y)\le 0$. But then $$\vert f_A(y)-f_A(x)\vert = f_A(x)-f_A(y) \ge f_A(x) = \frac1n,$$ so again $f_A$ must be discontinuous at $x$.

• Does something fail if $F_n$'s are not strictly increasing? Sep 28, 2011 at 23:05
• As closed sets stay closed under arbitrary intersections, you can take any set of $F_n$'s and make a strictly increasing set of $G_n$'s. First $G_n=\cap_{i=n}^{\infty}F_i$ is an increasing set. Then if any successive ones are the same, just delete the duplicates. Sep 28, 2011 at 23:08
• @LostInMath: If they’re not increasing, you can’t go from $x\notin F_n$ to $f_A(x)<1/n$. But the inclusions needn’t be strict; I’ll fix that. Sep 28, 2011 at 23:10
• @LostInMath: But you’re right that it’s not an oversight: it’s just a convenience for the argument that I chose. Sep 28, 2011 at 23:18
• Brian M. Scott Can you help Connection of complex $e^z$ and real Dirichlet please?
– BCLC
Aug 19, 2018 at 6:24