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Limit proof 
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#1
Nov2709, 09:22 AM

P: 74

I came across a problem in Calculus by Spivak, and I'm having trouble formalizing the proof.
Let [tex]A_{n}[/tex] bet a set of finite numbers in [0,1], and if [tex]m \neq n[/tex] then [tex]A_{n}[/tex] and [tex]A_{m}[/tex] are disjoint. Let f(x) be defined as f(x)=1/n if x is in [tex]A_{n}[/tex] and f(x)=0 if x is not in any [tex]A_{n}[/tex]. The question asks to prove the limit as x goes to a of f is 0 for any a in [0,1]. Now I thought: given an n, there are only finitely many elements of [tex]A_{n}[/tex] in a neighborhood of a. Choose the smallest such n, say [tex]n_{0}[/tex]. Then [tex]f(x)\leq1/n_{0}[/tex]. Restrict the neighborhood further so that none of the elements of [tex]A_{n_{0}}[/tex] are in the interval. Then choose the next n such that it's minimal. Obviously [tex]f(x)\leq1/n\leq1/n_{0}[/tex]. Successively doing this, for arbitrarily small x, f(x) will tend to 0. I don't know how to prove this using the limit definition; can someone help me out with what [tex]\delta[/tex] to choose? 


#2
Nov2709, 10:02 AM

P: 256

The smallest such n with what property? You're not very clear about that. Here's a proof: I'm going to assume f(a)=0; the other case is only a slight modification and should be instructive for you to think about. Fix [tex]\varepsilon>0[/tex], then there exists N > 0 with [tex]1/N < \varepsilon[/tex]. Let [tex]B_N = \cup_{ n = 1 }^N A_n[/tex], so B_N is a finite set, and [tex]a \not \in B_N[/tex] (why?). Now, let [tex]\delta > 0[/tex] be such that [tex]( a  \delta, a + \delta ) \cap B_N = \emptyset[/tex] (why does such a [tex]\delta[/tex] exist?). Then if [tex] a  x  < \delta[/tex], either [tex]x \in A_m[/tex] for m > N, or f(x) = 0. In either case, [tex]f(a)  f(x)<1/N<\varepsilon[/tex], as desired.



#3
Nov2709, 10:31 AM

P: 367

1) if a is in a set An 2) if a is not in a set An given an epsilon larger than 0, you must consider all f(x) = 1/n that fails ( larger than epsilon). Suppose you consider all 1/n/f(x) >= epsilon, then you can consider a union of sets (S) An such that f(x) >= epsilon for any element in a set An. Now you've reduced the problem to taking delta to be the min distance between the members of that set (S) and a (the distance between the greatest lower bound of (S) and a). If a is a member of the set itself, then you can use the exact same argument, since we are only concerned with behaviour as x APPROACHES a. Also, you know that you can always use this general argument because the interval between any numbers is always infinitely dense. 


#4
Nov2709, 12:40 PM

P: 74

Limit proof
Did you assume f(a)=0 because you had [tex]( a  \delta, a + \delta )[/tex]? Would [tex]0 <  a  x  < \delta[/tex] solve that? After all, since we are looking at the limit 0 then the expression [tex]f(a)  f(x)<\varepsilon[/tex] becomes [tex]f(x)  0=f(x)<\varepsilon[/tex], since the value of f at a doesn't matter. Ok I think I get it. I was having trouble formulating all this with the [tex]\delta  \varepsilon[/tex] definition. Thanks for all your help! 


#5
Nov2709, 01:05 PM

P: 367

The delta you are taking is the min ( ax : x [tex]\in[/tex] [tex]\cup[/tex] n=1 to i An ) Here is an intuitive version of the proof: A1A15A20A4<a>A7A9A10000A124921894381241 Any x's will be dispersed into different sets An, so take delta to be the min distance between a set and the point a. This interval between a and any x in sets An will always exist, since an interval of real numbers is always infinitely dense 


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