Prove a given limit using epsilon-delta

[Freye]

Homework Statement

Suppose that A_n is, for each natural number n, some finite set of numbers in [0,1] and that A_n and A_m have no members in common if m=/n. Define f as follows:

f(x) = 1/n if x is in A_n
f(x) = 0 if x is not in A_n for any n

Prove that the limit of f(x) as x goes to a is zero for all a in [0,1]

Homework Equations

for every $\epsilon$ there exists a $\delta$ such that
0<|x-a|<$\delta$ : |f(x)-L|<$\epsilon$

The Attempt at a Solution

Well I'm not even really sure how to start this one so maybe someone could just give me a hint for what to choose as my epsilon and hopefully I can take it from there.

 

[lippyka]

A:Let $\epsilon>0$. We want to prove that for all $a\in [0,1]$, there exists $\delta>0$ such that $$0<|x-a|<\delta\implies |f(x)-0|<\epsilon$$This is equivalent to finding a $\delta>0$ such that$$0<|x-a|<\delta\implies f(x)<\epsilon$$Let $n$ be the natural number such that $A_n$ is the closest set of numbers containing $a$. Then, $|x-a|<\frac{1}{2n}$ implies $x\in A_n$. Since, by hypothesis, $A_n$ and $A_m$ have no members in common if $m\neq n$, this implies that $f(x)=\frac{1}{n}<\epsilon$ if $|x-a|<\frac{1}{2n}$. Thus, if we set $\delta = \frac{1}{2n}$, then we have $$0<|x-a|<\delta\implies f(x)<\epsilon$$