# Set Limit

• November 3rd 2006, 06:46 AM
PauKelome
Set Limit
Could someone show me how to do the following proof?

Consider the set A such that A={p/(2^n) where p is a natural number, n is a natural number, and p/(2^n) is between 0 and 1 (not inclusive).

Let f(x)=0 if x is in A.
Let f(x)=1 if x is in [0,1] - A.

Prove that limit (as x approaches a) of f(x) does not exist when a is in [0,1].

Any help here would be appreciated.
• November 3rd 2006, 07:15 AM
Plato
Quote:

Originally Posted by PauKelome
Could someone show me how to do the following proof?

The answer to that is no.
Because the way it has been written it is not true!

Quote:

Originally Posted by PauKelome
Consider the set A such that A={p/(2^n) where p is a natural number, n is a natural number, and p/(2^n) is between 0 and 1 (not inclusive).
Let f(x)=0 if x is in A.
Let f(x)=1 if x is in [0,1] - A.
Prove that limit (as x approaches a) of f(x) does not exist when a is in [0,1].

If p=3 then $
A = \left\{ {\frac{3}{4},\frac{3}{8},\frac{3}{{16}},...} \right\}
$
now if $
a = \frac{8}{9}
$
then $
\lim _{x \to a} f(x) = 1.
$

In fact only for a=0 does the limit not exist.
• November 3rd 2006, 07:21 AM
CaptainBlack
Quote:

Originally Posted by PauKelome
Could someone show me how to do the following proof?

Consider the set A such that A={p/(2^n) where p is a natural number, n is a natural number, and p/(2^n) is between 0 and 1 (not inclusive).

Let f(x)=0 if x is in A.
Let f(x)=1 if x is in [0,1] - A.

Prove that limit (as x approaches a) of f(x) does not exist when a is in [0,1].

Any help here would be appreciated.

The basic idea here is that every neighbourhood of 0 contain a root of
f(x)=1, and also a root of f(x)=0.

In which case there is no k in R such that given epsilon >0, there is
beta>0, such that when 0<x<beta

|f(x)-k|< epsilon

because every inteval [0,beta) contains points at which f(x)=1 and f(x)=0.

RonL
• November 3rd 2006, 07:30 AM
Plato
Question: if this for any p and any n?
The way I read it, p is fixed?
• November 3rd 2006, 07:32 AM
ThePerfectHacker
Quote:

Originally Posted by Plato
The way I read it, p is fixed?

I think these are p-adic (in this case 2).
Thus, it can be anything.
• November 3rd 2006, 08:28 AM
PauKelome
p is not fixed. Apparently the set A is the set of all "dyadic rational numbers" between 0 and 1 if that makes it easier.
• November 3rd 2006, 08:30 AM
PauKelome
Does that clarification make it easier to prove?
• November 3rd 2006, 10:30 AM
Plato
Quote:

Originally Posted by PauKelome
Does that clarification make it easier to prove?

Easy is in the eye of the beholder. This can be hard not knowing the theorems you have to work with. There is one basic lemma that is important to this problem: Given any two numbers in [0,1] there is a dyadic rational number between them. Do you need to prove that? Also between any two numbers there is an irrational number. If $a \in [0,1]$ and $\varepsilon > 0$ then the neighborhood of a, $\left( {a - \varepsilon ,a + \varepsilon } \right)$ contains both a dyadic rational number and an irrational number in $\left[ {0,1} \right]$. Therefore, the values of $f$ at those numbers will differ be 1. So there can be no limit at $a$.