Need help with a limit problem on Spivak's Calculus.

[Elvz2593]

1. 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:
$f(x)=\left\{\begin{array}{cc}1/n,&\mbox{ if }<br /> x \in A{n} \\0, & \mbox{ if }x Not In A_{n} For Any n.\end{array}\right.$

Prove that $\lim$ x-a f(x)=0 for all a in [0,1].

Michael Spivak - Calculus Ch5 Q24

Homework Equations

The Attempt at a Solution

I've been trying to figure out the existence of this limit, but so far with no luck. I might have misunderstood the problem. Can someone help me out with this?Update: Just looked a older post on this problem. I think I overlooked some crucial stuff.
So, if we pick a delta value so small that none of the points in A1,A2...An are on the (a-delta, a+delta) interval, except maybe point a itself. f(x) approaches 0 to as x -> a, even if f(a) is defined at some 1/n. Am I being correct?

 

[spamiam]

Not quite. Remember, although there are only a finite number of elements in each $A_n$, there are an infinite number of the sets, one for each positive integer. So there's no "end" to the list of sets: $A_1, A_2, \ldots$. So I claim you may not be able to always find such an interval. For example, let $A_n = \{1/n\}$ for each $n \in \mathbb{Z}^+$. I claim that any neighborhood of 0 contains a point in $A_n$ for some n.

I think it might help to use the sequential version of functional limits.

EDIT: Oh, and welcome to PF!

 

[Elvz2593]

I think it is probably the author's wording error, otherwise it would be proved false.

Thank you so much for helping me out there. I like this place already.

 

[spamiam]

I think it is probably the author's wording error, otherwise it would be proved false.

No no, that wasn't a counterexample! As you'll note, we also have $\lim_{n \to \infty} 1/n = 0$, so the proposition holds in this case as well. (Spivak is a very good book, and I'm sure he wouldn't make such an error!) I just wanted to show that you can't always find an interval that doesn't hit a point in any of the A_n's.

Thank you so much for helping me out there. I like this place already.

Glad to hear it!

 

[Elvz2593]

I see what you are saying now. So, the proposition would still hold as 1/n is approaching 0 as well.

 

[spamiam]

After thinking about it some more, I think we can modify your original approach using cases. In the first case we can find a neighborhood of the point that doesn't contain any elements of the $A_n 's$ and we're done. Otherwise, what does that say about each neighborhood of a? What must each one contain?