If d>|t-c|>0 satisfies e>|f(t)-L| does d1<d also?

1. Dec 9, 2015

Odious Suspect

I was trying to come up with a proof of the well-known limit of products theorem. Suppose we have

$$\left|F_1 (t)-L_1\right|<\epsilon _1\Rightarrow \delta _1>|t-c|>0$$

and

$$\left|F_2 (t)-L_2\right|<\epsilon _2\Rightarrow \delta _2>|t-c|>0$$

$$\left|F_1 (t)-L_1\right|\left|F_2 (t)-L_2\right|<\epsilon _1\epsilon _2\Rightarrow min(\delta _1,\delta _2)>|t-c|>0$$

I could wave my hands and say that a well-behaved function approaches linearity in the neighborhood of $c$, or something that sounds similarly impressive, but I am not satisfied with that. It seems my method of proof requires that any $\delta < \delta _1$ will also satisfy $\left|F_1 (t)-L_1\right|<\epsilon _1$.

What justification, if any, have I for such an assertion? Again, this seems obvious for "well-behaved" functions, but I'm not sure which is the chicken and which is the egg.

I acknowledge that my proof seems a bit frumpy, even if it works; but the question about $\delta < \delta _1$ remains.

Last edited: Dec 9, 2015
2. Dec 9, 2015

Hi Odious:

Good luck.

Regards,
Buzz

3. Dec 9, 2015

Erland

You got this backwards. It should be $0 < |t-c| < \delta_1 \Rightarrow |F_1(t)-L_1|< \epsilon_1$.
That $\lim_{t\to c} F_1(t) = L_1$ means that to every $\epsilon_1 >0$ there is such a $\delta_1 >0$. Same for $F_2$.

This is also wrong. What you want to prove is that $\lim_{t\to c} F_1(t)F_2(t)=L_1L_2$, right? Then you must prove that to every $\epsilon > 0$ there is a $\delta>0$ such that $0<|t-c|<\delta\Rightarrow |F_1(t)F_2(t)-L_1L_1|<\epsilon$, and that is something completely different.

Yes, this is no problem. If $0 < |t-c| < \delta \Rightarrow |F(t)-L|< \epsilon$ and $0<\delta'<\delta$, then certainly $0<|x-c|<\delta'$ implies $0<|x-c|<\delta$, and hence $|F(t)-L|< \epsilon$. So if it works for some $\delta>0$, then it works for any smaller $\delta>0$. This simplifies things, since we don't need to find the "best" (in the sense "greatest possible") $\delta$, it suffices to find some $\delta$ which works.

4. Dec 9, 2015

Erland

5. Dec 11, 2015

Odious Suspect

That was just scratch-pad thinking, but I'm not sure it's wrong. The intent was: given $\epsilon_1$ the existence of some $\delta_1$ such that $0 < |t-c| < \delta_1$ is implied.

The following is from Fundamentals of Mathematics, Vol. 1: Foundations of Mathematics: The Real Number System and Algebra.
$$\underset{x}{\land }\underset{\epsilon }{\land }\left(x\in I\to \underset{\delta }{\lor }\underset{y}{\land }(y\in I\land |x-y|<\delta \to |f (x)-f (y)|<\epsilon )\right)$$

$\epsilon$ and $\delta$ are positive real numbers and I is a real number interval. I've followed their notation, rather than converting to the contemporary standard. It was given as an example of how to work with the symbols of propositional calculus. It is the Cauchy definition of continuity of a function $f$ in an interval $I$. I believe it addresses the essence of my original question, but I am not presenting it for that purpose. My point is that in English, I read it as: for all $x$ in $I$ and all $\epsilon$ it is implied that some $\delta$ exists for all $y$ in $I$ such that of $|x-y|<\delta$ it follows that $|f (x)-f (y)|<\epsilon$. So my $\Leftarrow$ was the first implication.

I hadn't spelled out my approach. That was just a starting point. I intended to expand the multiplication and come up with something equivalent to your statement. I have yet to return to the problem. I just wanted to explain myself a bit in this post.

No matter how much advanced math I read, I still get the sense that "it works because it has to work", is what "well-behaved function" means.

Q: What's a well behaved function?
A: A function that is differentiable.

Q: What is a differentiable function?
A: One that is well behaved.

6. Dec 11, 2015

Erland

But this "first implication" is not an implication, at least not in the mathematical sense. The existence of such a $\delta$, what is it implied by? Implication is a relation between propositions: $P\implies Q$, and here we have no propostion $P$. So instead of saying

"for all $x$ in $I$ and all $\epsilon$ it is implied that some $\delta$ exists for all $y$ in $I$ such that of $|x-y|<\delta$ it follows that $|f (x)-f (y)|<\epsilon$

one should say

"for all $x$ in $I$ and all $\epsilon>0$ there exists a $\delta>0$ such that for all $y$ in $I$: $|x-y|<\delta$ implies that $|f (x)-f (y)|<\epsilon$.

so that there is only one genuine implication here.

In your original post, you had an expression which was abbreviated in such a way that it looked like an implication facing the wrong direction. I understand that it was "scratchpad" writing, but you pubslished it here, and you can't expect us to be thought readers...

To be fair, there is in fact a way in which your mischaracterized implication can be interpreted as a genuine implication: we can interprete "For every $\epsilon>0$, there is a $\delta > 0$ such that..." as "For every $\epsilon$: If $\epsilon >0$, then there is a $\delta$ such that $\delta>0$ and...", but it quite rare to write out an implication like this.

I don't understand how this applies here.

(Native English speakers may very well find better formulations than mine. I am Swedish.)

7. Dec 11, 2015

Odious Suspect

8. Dec 11, 2015

Erland

Lack of time, Odious...? :) I think you meant this:

$\lim_{t\to c} f_1(t)=L1 \land \epsilon>0\implies \exists\delta>0:\forall x\in I (0<|t-c|<\delta \implies |f_1(t)-L1|<\epsilon)$ and this is a correct implication.

And, for your original question: If you find a $\delta$ satisfying this for a given $\epsilon >0$, the every smaller $\delta>0$ also satisfies this, for the same $\epsilon$. This is obvious. So if the task just is to find such a $\delta$, there is always infinitely many to choose from, and there is no need to take the greatest possible one, if it is simpler or more useful to take a smaller one.

9. Dec 12, 2015

Odious Suspect

Absolutely. Thank you for translating my babbling into a coherent statement. As I say, I need to work on my formal communication skills. My only reader for most of my scrawling is in the position to read my mind.

Well, there's obvious and then there's obvious IF.... For example, if I concoct a function $∀_{x∈ℝ}(∀_{x∈ℚ}(f(x)=sin(x)) \veebar f(x)=1)$. I believe that would allow me to always find $\delta$ satisfying the requirements of the definition of a limit, but it would not follow that $∀_{\delta>|x-c|>0}∀_{\delta>\delta^{'}>0} \implies \epsilon >|f(x)-L|$. That is to say, my pathological function defies the definition of a limit.

The reason for my original post was because I wanted to say things such as $\epsilon = \epsilon_1 = \epsilon_2$ or $\forall \delta_{'}:\delta>\delta_{'}>0 \implies \epsilon >|f(t)-L| \implies \delta_{'}>|t-c|$ when exhibiting proofs of theorems dealing with limits. In the former case, $\epsilon$ no longer seems truly "arbitrary", and the latter case seemed not to follow from the previously stated assumptions.

I'm not really sure how to state the necessary assumptions in a way appropriate for an introduction to calculus. And, in at least one case, the number theorists referred my to the topologists who referred me to the number theorists for the ultimate proof of the continuity of ℝ.