0

After seeing some proofs of epsilon delta x^2 , e^x, x^3, I notice that the common proof pattern is we take $|f(x)-f(c)|$ and find some expression $h(c,\delta): \mathbb{R} \times \mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0 }$ which is surjective (or better yet invertible) when seen as a function constant in $c$, so that we have:

$$|f(x)-f(c)| < h(c,\delta)$$

And we have a few cases on how it is proceeds now. If we didn't restrict $\delta$'s domain at any point in the proof (eg: as we do with $x^2$), then by surjectivity of $h$, for every $\epsilon \in \mathbb{R}_{\geq 0}$ we have a $\delta$ so that $\epsilon = h(\delta)$, now this is the $\delta$ required to satisfy the continuity criterion.

In case we can some how get an invertible $h$ we set $ \delta = h^{-1}(\epsilon)$. In cases of continuity of monomial terms, we also incoporate the trick of $\delta$ restriction to get such an invertible $h$, and then define $\delta$ through a minimum function to, so that the assignment of $\delta$ we get from epsilon is consistent with the restriction of $\delta$ we put before.

So, far so good, now I only have one question in the final thing which we get from our hard work of bounding, that is $h(c,\delta)$ why it in almost every case not $h(c,\delta,x)$. That is, why do we aim to get the final thing i.e $h$ , which is also what we equate to $\epsilon$, that the quantity $|f(x)-f(c)|$ is lesser than as independent of $x$?

I've been struggling with trying to answer this for my self. My weak intuition is this: $\delta$ directly restricts what values of $x$ can take up. If $h$ is somehow dependent of $x$, then so would $h^{-1}$ be too, in whatever case we are in. Now, it'd be weird if the quantity which tells us where $x$ is depends on $x$ itself... but then again recursive definitions do exist in mathematics... some maybe there are better reasons to give for this.

Note: I am very aware of the trivial answer "cause it works"

Clemens Bartholdy
  • 14,222
  • 1
    Because $x$ ranges over an interval of values. You have dropped the "$\forall x$" from your definition, and ignoring scope of variables, and interdependency of variables is a common mistake. – JonathanZ Aug 03 '24 at 18:26
  • https://math.stackexchange.com/questions/2686159/why-cant-delta-depend-on-x-in-delta-epsilon-proofs – Clemens Bartholdy Aug 04 '24 at 07:59

1 Answers1

4

This comes down to the correct interpretation of the $\epsilon$-$\delta$ definition of limit (or of continuity). The definition looks like this:

For all $\epsilon > 0$, there exists $\delta > 0$ such that for all $x$, ...

When you have a sequence of alternating quantifiers like this (for all, there exists, for all, ...), the ordering is critical. The interpretation of such a statement is that each variable is allowed to depend on the previously mentioned variables, but not the later ones. Here they are in the order $\epsilon$, $\delta$, $x$. So $\delta$ is allowed to depend on $\epsilon$, but not on $x$.

Consider this simpler case where we change the order of quantifiers:

(1) For all real numbers $x$, there exists a real number $y$ such that $y > x$.

(2) There exists a real number $y$ such that for all real numbers $x$, we have $y > x$.

In (1), $y$ is allowed to depend on $x$, but in (2), $y$ is not allowed to depend on $x$. This makes a big difference: (1) is true and (2) is false.

Ted
  • 35,732