What's the difference between these two statements??

• Jul 8th 2010, 08:08 PM
crabchef
What's the difference between these two statements??
This question is driving me nuts. I've gotten plenty of hints from office hours and emails, but I still don't quite understand it!

The questions is, of the following two statements, which implies which? Basically, which statement is stronger? The second part of the question is, does there exist a function for which both statements are true?

(a) $\displaystyle \forall \varepsilon>0, (\exists \delta>0,(\forall a\in \mathbb{R}, if |x-a|<\delta, then |f(x)-f(a)|<\varepsilon))$

(b) $\displaystyle \forall \varepsilon>0, (\forall a\in \mathbb{R}, (\exists \delta>0 if |x-a|<\delta, then |f(x)-f(a)|<\varepsilon))$

---------------

The help that I have gotten for this problem is:

PROBLEM 2.26: You are asked to two things:
1. Tell which statement implies the other.
(Does (a) imply (b), or does (b) imply (a)?)
2. Find a function where both are true or explain why it is impossible for
both to be true.

The first thing I would is write out the statments paying attention to quantifiers, the order of quantitiers and looking for differences between the statements. Here I have translated each statement into our standard form:

a) FOR ALL epsilon > 0,
FOR ALL real numbers a,
THERE EXISTS delta > 0 such that
if |x-a| < delta, then |f(x) - f(a)| < epsilon.

b) FOR ALL epsilon > 0,
THERE EXISTS delta > 0 such that
FOR ALL real numbers a,
if |x-a| < delta, then |f(x) - f(a)| < epsilon.

Question 1 asks: If you are given a function for which (a) is TRUE,
then is it true that (b) would also be TRUE for that function?
Or in the other direction, if you are given a function for which (b) is TRUE,
then is it true that (a) would also be TRUE for that function?

Another way to say this is: which one is more specific and which one is more general? The more specific statement will also satisfy the more general statement. This all depends on the order of the quantifiers.

Question 2 asks: Can you find a function for which both are true?
The actual "if ... then ..." part of the statment is saying that
if |x-a| < delta, then |f(x) - f(a)| < epsilon
which means
if x is within a distance of delta from a,
then the output f(x) is within a distance of eplison from f(a).
In other words, this is a statement about functions where the inputs
being close together gaurantee the outputs are close together.

-------------------------

When I went into office hours, he showed us that f(x)=x^2 works for the first statement but fails for the second statement whereas f(x)=x works for both statements.

I don't really know what this means. I don't understand how this problem is a "statement about functions where the inputs being close together guarantee the outputs being close together."

Thank you for any help, or even just reading the problem hhaha
• Jul 8th 2010, 08:59 PM
undefined
Quote:

Originally Posted by crabchef
When I went into office hours, he showed us that f(x)=x^2 works for the first statement but fails for the second statement whereas f(x)=x works for both statements.

I might have made a mistake, but I think f(x)=x^2 works for the second statement but fails for the first. Reasoning below.

Quote:

Originally Posted by crabchef
This question is driving me nuts. I've gotten plenty of hints from office hours and emails, but I still don't quite understand it!

The questions is, of the following two statements, which implies which? Basically, which statement is stronger? The second part of the question is, does there exist a function for which both statements are true?

(a) $\displaystyle \forall \varepsilon>0, (\exists \delta>0,(\forall a\in \mathbb{R}, if |x-a|<\delta, then |f(x)-f(a)|<\varepsilon))$

(b) $\displaystyle \forall \varepsilon>0, (\forall a\in \mathbb{R}, (\exists \delta>0 if |x-a|<\delta, then |f(x)-f(a)|<\varepsilon))$

Statement (b) means that for all real numbers $\displaystyle \displaystyle a$, the limit $\displaystyle \displaystyle \lim_{x \to a}f(x)$ exists and equals f(a). Statement (a) on the other hand means that there is a way to choose $\displaystyle \displaystyle \delta$ that relies only on $\displaystyle \displaystyle \epsilon$ and not on $\displaystyle \displaystyle a$, to satisfy the definition of limit/continuity for all real $\displaystyle \displaystyle a$. (Recall that many epsilon-delta proofs from calculus involve finding an expression for delta in terms of a and epsilon. See here for an example.) Can you see why (a) is stronger and implies (b)?

If I've made a mistake, I'm pretty sure someone will come along and correct me before long.
• Jul 8th 2010, 09:13 PM
crabchef
Thanks for taking a crack at the problem. I really don't understand why (a) is stronger and implies (b). The professor said that the first one is something called continuous and the second is something called uniformly continuous. Being an awful math person, while I understand what those mean, I have the feeling that I don't even understand what this problem is really talking about.
• Jul 8th 2010, 09:13 PM
MattMan
Perhaps this will help

Uniformly_continuous

What you have is a. the definition of uniform continuity and b. the definition of just continuity.
So in a, there exists a $\displaystyle \delta$ dependent on $\displaystyle \epsilon$ only and independent of what a is.
In b, there exists a $\displaystyle \delta$ dependent on $\displaystyle \epsilon$ and a.
So if we have $\displaystyle \delta (\epsilon)$ then we clearly satisfy $\displaystyle \delta(\epsilon,a)$ (if we are thinking about delta as a function of the other variables)
• Jul 11th 2010, 07:41 AM
HallsofIvy
What is the difference between "a cup of sifted flour" and "a cup of flour, sifted"!(Giggle)