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

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

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?
 
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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!
 
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.:biggrin:
 
Elvz2593 said:
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 An's.

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

Glad to hear it!
 
I see what you are saying now. So, the proposition would still hold as 1/n is approaching 0 as well.
 
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 &#039;s and we're done. Otherwise, what does that say about each neighborhood of a? What must each one contain?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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