Spivak calculus ch.5 #24 delta epsilon proof for limit of peicewise function

[balingwhale]

Homework Statement

Suppose A_n is, for all natural numbers n, some finite set of numbers in [0,1] and A_n intersect A_m={ } if m!=n

Define f as follows:

f(x) = 1/n if x is in A_n and 0 if x is not in A_n for all n.

Prove that the limit as x goes to a of f(x) = 0 for all a in [0,1].

Homework Equations

The Attempt at a Solution

a is in [0,1]
x is in (a-delta, a+delta)
suppose a is in some A_n (a finite set) then how do we show that x near a is not also in A_n and can A_n be equal to set of all numbers in the closed interval [0,1]?

this problem is very confusing to me, where do i start?

 

[jgens]

Let \varepsilon > 0 and choose n so large that \varepsilon \geq 1/n. If these conditions hold, there are only finitely many numbers such that |f(x) - 0| < \varepsilon can fail. Start here and work onwards.

 

[balingwhale]

Let \varepsilon > 0 and choose n so large that \varepsilon \geq 1/n. If these conditions hold, there are only finitely many numbers such that |f(x) - 0| < \varepsilon can fail. Start here and work onwards.

thank you for your quick reply.
to clarify, \varepsilon \geq 1/n? not \varepsilon \leq 1/n?
further,
should i suppose that a is in A_n?
then either x near a is also in A_n and the limit = 1/n = 0 since n is large?
or x near a is not also in A_n and the limit = 0.

when will the inequality fail?
also, can A_n = the set of all numbers in the closed interval [0,1] ?

 

[LCKurtz]

So f(x) = 1/n if x is in A_n and f(x) = 0 is x is not in any A_n. So there are finitely many points (those in A₁) where f(x) = 1, finitely many others (those in A₂) where x is 1/2, etc. So as you go to larger subscripts the values of f(x) for x in A_n get small. And those x not in any A_n have f(x) = 0. So once you get past A_n (passing only finitely many terms in the process) f(x) is small or zero.

Does that help?

 

[balingwhale]

why should you get past A_n? what if a is in A_1?

 

[LCKurtz]

What if it is, so what? Can you pick delta small enough that none of the finitely many "big" values are in the delta interval around a? Remember, we're not talking about f(a) here, but the limit as x nears a.

 

[emyt]

why should you get past A_n? what if a is in A_1?

if a is in A_1, how will it approach a? can a set An have all possible values between x and a?

 

[emyt]

when will the inequality fail?
also, can A_n = the set of all numbers in the closed interval [0,1] ?

no, there is an infinite amount of numbers between 0 and 1. An is a finite set

 

[LCKurtz]

Each A_n is a finite set. You are trying to show that the limit as x --> a is 0. So it's the x nearby to a for which you need f(x) to be small.

 

[balingwhale]

What if it is, so what? Can you pick delta small enough that none of the finitely many "big" values are in the delta interval around a? Remember, we're not talking about f(a) here, but the limit as x nears a.

what do you mean by the finitely many "big" values in the delta interval around a?

so we're trying to say that x near a is never in the same set as a? and if we pick n to be very large then the sets become smaller?

choose epsilon >= 1/n as before, s.t |x-a|< delta => |f(x)-0|=|f(x)|< 1/n

the epsilon inequality will fail if x (within delta of some a that can be in any set) is in one of A_1, A_2, ..., A_n? these are finitely many n sets followed by infinitely many more sets that might contain x? still in [0,1]?
then what? there will always exist an x that is even closer to a that is not in any of A_1, .. A_n ?

 

[emyt]

so we're trying to say that x near a is never in the same set as a?

x -> a has an infinite amount of terms, if you had 100 terms near a in the same set, youll still have a an infinite amount of terms outside of that set.

 

[jgens]

Given the OP's struggle and since the he/she is working out of Spivak's calculus book, I would recommend revisiting his proof that \lim_{x \to a}f(x) = 0 where the function f is defined by f(x) = 1/q for rational x in lowest terms and f(x) = 0 for irrational x. This proof is almost a restatement of this argument with a few minor changes.

 

[balingwhale]

Given the OP's struggle and since the he/she is working out of Spivak's calculus book, I would recommend revisiting his proof that \lim_{x \to a}f(x) = 0 where the function f is defined by f(x) = 1/q for rational x in lowest terms and f(x) = 0 for irrational x. This proof is almost a restatement of this argument with a few minor changes.

thanks a bunch but i think i had it in my last post.
there will always be an x' that is closer to a than x and is not contained within one of the finite sets A_n.

 

[LCKurtz]

It doesn't look to me like you have it yet. Given a point a in [0,1] and an ε > 0, you need to explain how you can find a δ > 0 such that |f(x)| < ε whenever 0 < |x - a| < δ. Until you can explain how to find that δ you haven't understood the problem.