Supose that for any natural number $n$, $A_n$ is a finite set of numbers from $[0,1]$, and that $A_m$ and $A_n$ have no common elements if $m \neq n$, ie
$$m \neq n \Rightarrow A_n\cap A_m=\emptyset$$
Let $f$
$$f(x)= \begin{cases} 1/n & \text{for } x \in A_n \cr 0 & \text{for } x \notin A_n \text{ for any }n \end{cases}$$
I guess the definition is clear: If $x$ is in some of the $A_n$ then we map it to $1/n$, and if $x$ is in no $A_n$ the function is zero. It's like a modified characteristic function.
$$f(x) =\sum_{n \in \Bbb N} \frac 1 n \chi_{A_n}$$
(Thanks Asaf)
I have to prove that $$\lim_{x \to a }f(x)=0$$ for all $a$ in $[0,1]$
This is my inutuitive interpretation of the problem.
Since all the $A_n$ are finite sets, the union $S= \bigcup_{n \in \Bbb N}A_n$ of the sets is countably infinite. (Maybe this has to be proven before, but I think it is true.)
This means that $f(x)\neq 0$ for countable infinite many $x$. But then the set of $x$ such that $f(x)=0$ is uncountable, since $[0,1]$ is uncountable so $f$ is $0$ almost everywhere in $[0,1]$.
Although this is not homework, I'd like you to help me find the way to the "solving argument", maybe show how it can be done for $a=1/2$. This is from Spivak's Calculus, so all the set theoretic things I wrote don't really apply, it should probably be proven by some basic set arguments and the definition of the limit.
Attempt of proof:
DEFINITION: $$\lim_{x \to a}f(x)=L$$ if $\forall \epsilon >0 \exists \delta >0 : 0<|x-a|<\delta \Rightarrow |f(x)-L|<\epsilon$. This is Spivak's definition. Note that
$0<|x-a|$ means the limit excludes the point $a$. Suggested by t.b. is the notation
$$\lim_{\substack{x \to a \\ x \neq a}} f(x)$$
T Let $f : [0,1] \to \mathbb Q$ such that $$f(x) = \sum_{n \in \Bbb N} \frac{\chi_{A_n}(x)}{n}$$
then $$a \in [0,1] \Rightarrow \lim_{x \to a} f(x) = 0$$
P (Based on Zhang's idea).
Since $A_1$ is finite, there exists a $\delta_1 >0$ such that no $x \in A_1$ is in $(a-\delta_1,a)\cup (a,a+\delta_1)$. Thus, $|f(x)|< \dfrac 1 2$ for $0<|x-a|<\delta_1$. Similarily, $\exists \delta_2 >0 : x\in A_2 \wedge x \notin (a-\delta_2,a)\cup (a,a+\delta_2) $ so $|f(x)|< \dfrac 1 3 $ for $0<|x-a|<\delta_2$. Analogously,
$$\exists \delta_n >0 : x\in A_n \wedge x \notin (a-\delta_n,a)\cup (a,a+\delta_n) $$, so $$|f(x)|< \dfrac 1 n \text{ for } 0<|x-a|<\delta_n$$
Revised:
Let $\epsilon>0$ be given. Let $N\in \mathbb N$ such that $1/N < \epsilon$. Let $n \geq N$, and $$\delta = \min \{ \delta_1,\cdots,\delta_n\}$$ Then
$$0<|x-a|<\delta \Rightarrow |f(x)|<\epsilon$$ ∎.
Today I was discussing this problem with a professor and at first glance he thought it was the case that
$$S= \bigcup_{n \in \Bbb N}A_n= [0,1]$$
I provided the following explanation.
By Cantor's proof, $[0,1]$ is uncountable. I'll use $\sim$ to say there is a bijection between two sets $A$ and $B$. Since all the $A_n$ are finite, their cardinality is a natural number, so
$$A_1 \sim \left \{ 1,2,\cdots, |A_1| \right \}$$
$$A_2 \sim \{ |A_1|+1,\cdots, |A_1|+|A_2| \}$$
$$\cdots$$
$$A_n \sim \left\{ \sum_{k <n}|A_k|+1,\cdots, \sum_{k \leq n} |A_k|\right\}$$
Then, taking the union produces
$$\bigcup_{n \in \Bbb N}A_n\sim \Bbb N$$
so the set $S$ is countable and thus can't be $[0,1]$.
So far, I have these satisfactory ideas, but I want to write an acceptable proof.
anon: If $g$ and $f$ differ at only a finite number of points then $\lim f = \lim g$. We define a useful $g_m$ such that it differs with $f$ at only a finite amount of points, and we show $0 \leq g_m \leq 1/m$.
I got this one and hope I can devise a proof.
Levon/Glouglou/Zhang Show that for any sequence $x_n$ s.t. $x_n \to a$, there exists an $n_0$ such that $$\{ x_k \}_{k \geq n_0}\cap A_n=\emptyset$$ for all $n \in \Bbb N$.
This means that for a suitable $\delta$, the set $M=\{ x : x \in [a-\delta,a+\delta]\}$ contains no $x \in A_n$, so $f(x) < \epsilon$ (actually it is strictly $0$) in that neighborhood of $a$.
Best Answer
Your interpretation is not quite correct. For example the function $$f(x) = \left\{ \begin {array} {ll} 1 & x \text { is rational} \\ 0 & x \text { is irrational} \end {array} \right.$$ also has the property that it is nonzero only on countably many points. However, at no point the limit of this function exists. (Because you can approximate each number by rationals and by irrationals).
As a hint I would suggest the following. If $a, \varepsilon$ and $n$ are given, then you can peak $\delta$ so small that the $\delta$ neighbourhood of $a$ does not contain numbers from $A_1, ..., A_n$.