- #1

- 23

- 7

## 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]##.##f(x) = 0##, if ##x \notin A_n## for all ##n##

## 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##.

So ask somebody to rip apart your proof completely.

*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##.