# Prove that function tends to 0 everywhere in this interval

## Homework Statement

From Spivak: 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\neq n##. Define f as follows:
##f(x) = \frac{1}{n}##, if ##x \in A_n##
##f(x) = 0##, if ##x \notin A_n## for all ##n##​
Prove that ##\lim_{x\to a} f(x) = 0## for all ##a## in ##[0,1]##.

## Homework Equations

Epsilon-Delta definition of a limit

## The Attempt at a Solution

I'm trying to follow micromass' suggestion, so I would be delighted if someone could assist me. My main concern is that I did not really make use of the assumption that ##A_n## and ##A_m## have no members in common if ##m\neq n##.
Proof: We first prove the limit tends towards ##0## for all ##a## in the open interval ##(0,1)##. So let ##\epsilon >0## be arbitrary, and let ##a## be an arbirtrary element of ##(0,1)##. We can find some natural number ##n_{\epsilon}## such that for all ##m\in \mathbb{N}## for which ##m\geq n_{\epsilon}##, ##\frac{1}{m} < \epsilon##, so there is only a finite number of natural numbers ##1,...,n_{\epsilon}-1## whose reciprocals are greater than ##\epsilon##, and therefore finitely many ##y\in [0,1]## for which ##|f(y)| > \epsilon##. Hence, we will be able to find a neighbourhood of ##a## which does not contain any of the ##y##'s, and hence there exists a ##\delta## such that for all ##x##, ##0<|x-a|<\delta## implies ##|f(x)|<\epsilon##. For the endpoint ##0##and ##1##, the right and left hand limits both tend to 0 respectively, and since ##f(x) = 0## for all ##x \notin [0,1]##, we can conclude that ##\lim_{x\to 0}f(x) = \lim_{x\to 1}f(x) = 0##.

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Samy_A
Homework Helper
My main concern is that I did not really make use of the assumption that ##A_n## and ##A_m## have no members in common if ##m\neq n##.
That assumption is (implicitly) used in the definition of the function ##f##. If ##A_n## and ##A_m## have members in common, ##f## is not well-defined.

lordianed
If AnA_n and AmA_m have members in common, ff is not well-defined.
Oh right, thanks Samy, I did not catch that

Hi lordianed ! This exercise is very much like the previous one you put on PF a few days ago.

Assume that your function converges to a limit ##\ell## as ##x\rightarrow a##. Assume that ## a \in ]0,1[## and adapt the definition for ##a\in\{0,1\}##.

So you have,

##\forall \epsilon > 0,\ \exists\delta >0\ \forall x\in [0,1] \ ( x\in[ a - \delta , a + \delta ] \Rightarrow |f(x) - l | \le \epsilon) ##

You need to show that in ##[ a - \delta , a + \delta ]##, there is an element ##x_0## that belongs to none of the ##A_n##. In which case ## |\ell| = |f(x_0) - \ell | \le \epsilon##. That proves that if the limit exists, it has to be zero.

Hi lordianed ! This exercise is very much like the previous one you put on PF a few days ago.

Assume that your function converges to a limit ##\ell## as ##x\rightarrow a##. Assume that ## a \in ]0,1[## and adapt the definition for ##a\in\{0,1\}##.

So you have,

##\forall \epsilon > 0,\ \exists\delta >0\ \forall x\in [0,1] \ ( x\in[ a - \delta , a + \delta ] \Rightarrow |f(x) - l | \le \epsilon) ##

You need to show that in ##[ a - \delta , a + \delta ]##, there is an element ##x_0## that belongs to none of the ##A_n##. In which case ## |\ell| = |f(x_0) - \ell | \le \epsilon##. That proves that if the limit exists, it has to be zero.
Thanks Geoffrey, I'll try that method too. However I asked to have my proof ripped apart so to say, I already have tried proving it and would love to see people point out any flaws in my argument in the original post.