# Limit proof

by Bleys
Tags: limit, proof
 P: 74 I came across a problem in Calculus by Spivak, and I'm having trouble formalizing the proof. Let $$A_{n}$$ bet a set of finite numbers in [0,1], and if $$m \neq n$$ then $$A_{n}$$ and $$A_{m}$$ are disjoint. Let f(x) be defined as f(x)=1/n if x is in $$A_{n}$$ and f(x)=0 if x is not in any $$A_{n}$$. 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 $$A_{n}$$ in a neighborhood of a. Choose the smallest such n, say $$n_{0}$$. Then $$f(x)\leq1/n_{0}$$. Restrict the neighborhood further so that none of the elements of $$A_{n_{0}}$$ are in the interval. Then choose the next n such that it's minimal. Obviously $$f(x)\leq1/n\leq1/n_{0}$$. 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 $$\delta$$ to choose?
 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 $$\varepsilon>0$$, then there exists N > 0 with $$1/N < \varepsilon$$. Let $$B_N = \cup_{ n = 1 }^N A_n$$, so B_N is a finite set, and $$a \not \in B_N$$ (why?). Now, let $$\delta > 0$$ be such that $$( a - \delta, a + \delta ) \cap B_N = \emptyset$$ (why does such a $$\delta$$ exist?). Then if $$| a - x | < \delta$$, either $$x \in A_m$$ for m > N, or f(x) = 0. In either case, $$|f(a) - f(x)|<1/N<\varepsilon$$, as desired.
P: 367
 Quote by Bleys I came across a problem in Calculus by Spivak, and I'm having trouble formalizing the proof. Let $$A_{n}$$ bet a set of finite numbers in [0,1], and if $$m \neq n$$ then $$A_{n}$$ and $$A_{m}$$ are disjoint. Let f(x) be defined as f(x)=1/n if x is in $$A_{n}$$ and f(x)=0 if x is not in any $$A_{n}$$. 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 $$A_{n}$$ in a neighborhood of a. Choose the smallest such n, say $$n_{0}$$. Then $$f(x)\leq1/n_{0}$$. Restrict the neighborhood further so that none of the elements of $$A_{n_{0}}$$ are in the interval. Then choose the next n such that it's minimal. Obviously $$f(x)\leq1/n\leq1/n_{0}$$. 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 $$\delta$$ to choose?
you're on the right track, you should break this down into 2 parts:
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.

P: 74
Limit proof

 The smallest such n with what property? You're not very clear about that
Sorry, what I meant to say was the smallest n such that the set $$A_{n}$$ has an element in the neighborhood of a.
 so B_N is a finite set, and (why?)
f(a)=0 iff a=0, by definition of f, and this is iff a is not in any An.
 (why does such a $$\delta$$ exist?)
Because B_N is finite.

Did you assume f(a)=0 because you had $$( a - \delta, a + \delta )$$? Would $$0 < | a - x | < \delta$$ solve that? After all, since we are looking at the limit 0 then the expression $$|f(a) - f(x)|<\varepsilon$$ becomes $$|f(x) - 0|=|f(x)|<\varepsilon$$, since the value of f at a doesn't matter.

 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
So since S will be finite (since there will be an n such that $$1/n<\varepsilon$$) then that delta will work. Can I just not consider whether a is in the set or not. After all like you said, we're interested in the limit, not continuity. So $$0 < | a - x | < \delta$$ would solve that.
Ok I think I get it. I was having trouble formulating all this with the $$\delta - \varepsilon$$ definition. Thanks for all your help!
P: 367
 Quote by Bleys Sorry, what I meant to say was the smallest n such that the set $$A_{n}$$ has an element in the neighborhood of a. f(a)=0 iff a=0, by definition of f, and this is iff a is not in any An. Because B_N is finite. Did you assume f(a)=0 because you had $$( a - \delta, a + \delta )$$? Would $$0 < | a - x | < \delta$$ solve that? After all, since we are looking at the limit 0 then the expression $$|f(a) - f(x)|<\varepsilon$$ becomes $$|f(x) - 0|=|f(x)|<\varepsilon$$, since the value of f at a doesn't matter. So since S will be finite (since there will be an n such that $$1/n<\varepsilon$$) then that delta will work. Can I just not consider whether a is in the set or not. After all like you said, we're interested in the limit, not continuity. So $$0 < | a - x | < \delta$$ would solve that. Ok I think I get it. I was having trouble formulating all this with the $$\delta - \varepsilon$$ definition. Thanks for all your help!
It's a good idea to say how the limit works in both cases, you need to show that you've considered that case (it is a significant case). But in the end, it uses the same argument.
The delta you are taking is the min ( |a-x| : x $$\in$$ $$\cup$$ n=1 to i An )

Here is an intuitive version of the proof:

|A1||A15||A20||A4|<----a------>|A7||A9||A10000||A124921894381241|

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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