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In every epsilon delta proof I see, it ends up with reverse engineering the epsilon inequality to find a suitable delta. Ultimately, it seems so that one can always write down $\delta$ as a function of $\epsilon$. What I want to know is, does there exist any epsilon delta proof, in which we can only show that the delta only exists (as is the minimum condition in the statement) rather than the explicit relation of it with epsilon?

I am of opinion that there may exist no such proof. If the answer to the previous question is there exists no known ones, why not just keep it in the statement of epsilon delta limit that "$\delta$ can be written as some function of $\epsilon $ , $\delta=f(\epsilon)$ with $f: \mathbb{R^+} \to \mathbb{R^+}$ "

  • Your $f$ will depend on the particular question you are trying to answer. In a sense the existence of such an $f$ is much the same thing as the existence of a suitable $\delta$ for each $\epsilon>0$ – Henry May 15 '22 at 13:07
  • no. In the statement for epsilon delta, it says we only need existence of delta. In every single thing I saw, we get some function relating $\delta(\epsilon)$. I am asking, if there is any such problem in which we can only show such an existence rather than some function @AdamRubinson – tryst with freedom May 15 '22 at 13:09
  • The existence of a $\delta$ for each $\epsilon$ is precisely a function $\delta$ that for each $\epsilon$ spits out a particular $\delta(\epsilon)$. And people do write $\delta_\epsilon$, particularly when trying to keep track of dependencies – Calvin Khor May 15 '22 at 13:15
  • Ok, so it's just finding a function. Why does no book write it like this... it seems much more clear what we have to do when writing a proof this way... – tryst with freedom May 15 '22 at 13:17
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    It's always the case that $\delta = f(\varepsilon)$ for some function $f : \Bbb{R}^+ \to \Bbb{R}^+$ (this may depend on axiom of choice). It may be that, in certain difficult (and probably fairly artificial) cases, this function may be uncomputable, and thus impossible to "write down". That might be the closest you're going to get to an answer. – Theo Bendit May 15 '22 at 13:18
  • On the other end of the scale, constant functions don't need $\delta$ to be any function of $\varepsilon$. All you need to show that constant functions are continuous is the fact that at least one positive number exists (well, even if positive numbers didn't exist, that would mean everything would be vacuously continuous, due to a lack of $\varepsilon > 0$). – Theo Bendit May 15 '22 at 13:19
  • I do not understand what you mean when you say the function is uncomputable. Could you explain that more? also where did the AOC thing come from – tryst with freedom May 15 '22 at 13:20
  • @Aplateofmomos I mean, no Turing machine is capable of taking $\varepsilon$ and producing the corresponding $\delta$. There are constant functions like this, e.g. a function that constantly produces Chaitin's constant. – Theo Bendit May 15 '22 at 13:22
  • I thought computability related to algorithims. You had said functions. Aren't those two different things..? – tryst with freedom May 15 '22 at 13:24
  • Incomputable basically means "very difficult to compute". There are many things in number theory which are hard to compute. I have seen statements like, "Here is a proof that there exists a prime number between $10^{50}$ and $10^{180}$ that satisfies property $A$, thus proving theorem $7.3$." However, they have not been able to narrow down exactly where this number is in this interval... – Adam Rubinson May 15 '22 at 13:25
  • The fact that it is a function is by itself no more informative than just having it depend on $\epsilon$. There is the opposite scenario where we know that we can choose $\delta = f(\epsilon)$ for a well behaved $f$, and this is useful: eg Lipschitz functions satisfy $|f(x)-f(y)|\le C|x-y|$ for some $C>0$. – Calvin Khor May 15 '22 at 13:26
  • Doesn't a computer computing essentially mean to execute some sort of algorithim? – tryst with freedom May 15 '22 at 13:26
  • Well yes, computing usually does require algorithms, but what is your point... – Adam Rubinson May 15 '22 at 13:27
  • So how can a function be uncomputable? Is a function an algorithim...? – tryst with freedom May 15 '22 at 13:28
  • In certain areas in analysis, an explicit term for $\delta$ in terms of $\epsilon$ is impossible and uninteresting to describe explicitly. This happens when dealing with abstract cases. – FShrike May 15 '22 at 13:32
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    Consider the proof of the fact if $f:B\rightarrow C$ and $g:A\rightarrow B$ are continuous, then $f\circ g$ is continuous. To proof this we cannot use any explicit relations between $\epsilon$ and $\delta$, since $f$ and $g$ are not explicitly given. – Leander Tilsted Kristensen May 15 '22 at 13:32
  • @LeanderTilstedKristensen although if you are explicitly given two function $f$ and $g$ then you can find an explicit relationship between $\delta$ and $\varepsilon.$ – Adam Rubinson May 15 '22 at 13:34
  • @LeanderTilstedKristensen Aren't you? You're given arbitrary $f,g$, sure, and thus you can't actually write down any formulas. But they are still explicit. $f$ and $g$ both presumably carry their own "functional" relationship between $\delta$ and $\epsilon$, and you use that to explicitly construct a relationship for the composed function. – Arthur May 15 '22 at 13:47
  • @Arthur Point taken. I guess that one has to consider the exact meaning of the word "explicit". For instance if you rely on the axiom of choice to explicitly define something, is it then explicitly defined? (this is meant as an open question not a retorical one) – Leander Tilsted Kristensen May 15 '22 at 14:11
  • Why do you need axiom of choice for explicit definitiosn??? @LeanderTilstedKristensen – tryst with freedom May 15 '22 at 16:26
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    @Aplateofmomos To define $\delta(\epsilon)$ as a function of $\epsilon$, you must for each $\epsilon$ make a choice of what you want $\delta(\epsilon)$ to represent. This requires you to make an infinite number of choices, which is made possible by the axiom of choice. I do however not believe that the axiom of choice is necessary, for instance you could define $$\delta(\epsilon) := \frac{ \sup {\delta > 0 : | : |x-a| < \delta \Rightarrow |f(x)-f(a)|< \epsilon}}{2}$$ if the supremum is finite and $\delta(\epsilon) = 1$ otherwise. – Leander Tilsted Kristensen May 15 '22 at 17:47

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I think there is no real conceptual difference between the existence of a $\delta_\epsilon$-function and the standard: For all $\epsilon$ there exists $\delta>0$ s.t...

There is, however, a subtle difference between 'you may find' / 'there exists'.

In a typical phrase regarding $\epsilon,\delta$ it reads something like: Given any $\epsilon>0$ 'you may find' / 'there exists' $\delta>0$ such that ...

But there is also usually a monotonicity issue behind the curtain: If $\delta_0>0$ works for a given $\epsilon_0>0$ then $\delta_0$ also works for any $\epsilon>\epsilon_0$ and similarly any $0<\delta<\delta_0$ would work for the value $\epsilon_0$.

Assuming this monotonicity, to construct a $\delta(\epsilon)$ function you need induction but not AoC. It suffices to look at $\epsilon_k=1/k$, $k\geq 1$ and get ('find'/'there exists') a corresponding $\delta'_k>0$. Let $\delta_k=\min\{\delta'_1,\cdots,\delta'_k\}>0$. Then $\delta(\epsilon)=\max\{\delta_k : k>1/\epsilon\}$ should do.

So the equivalent statement for this $\delta$ function would be: 'You may find' / 'There exists' a monotone decreasing function $\delta : \epsilon \in (0,+\infty) \to \delta_\epsilon\in (0,+\infty)$ with $\lim_{\epsilon\to 0^+} \delta_\epsilon=0$ such that...

Thus, I think the real underlying question persists also when thinking in terms of such a function. Can you actually construct such function, or do you only know it exists?

For example: Perhaps it is consistent with ZF to assume that for every continuous function on the reals one may construct such a $\delta$-function and then all standard analysis (without AoC) works. Perhaps within ZFC you could construct an abstract function for which it is not possible to construct such a $\delta$ function? Frankly, I have no idea as to the answer of these two questions.

H. H. Rugh
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When proving that every real polynomial of a single variable is continuous as a function $\mathbf R \to \mathbf R$, you induct on the degree of the polynomial and rely on continuity of addition and multiplication as mappings $\mathbf R^2 \to \mathbf R$, but you don't rely on an "explicit relation" of $\delta$ in terms of $\varepsilon$ since that would be a big mess and quite distracting from the real ideas in the proof.

KCd
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  • Are you sure that the induction proof your referring to is non-constructive? Because I strongly suspect there is a concrete $\delta(\epsilon)$ function lurking just behind the curtains of that proof, it just isn't written down explicitly in its full form. – Arthur May 15 '22 at 13:45
  • "it just isn't written down explicitly in its full form" -- well, the original question asked if there are proofs where $\delta$ is not written down explicitly. That is such a proof. I understand what you're getting at, but my point was that in the proof of the result I mentioned, nobody in their right mind would try to write down an explicit expression for $\delta$ in terms of $\varepsilon$. I read the original post as being about "explicit/not explicit" rather than "constructive/non-constructive". If something remains behind the curtains, it is not explicit. – KCd May 15 '22 at 14:06
  • I mean here, it could be bought of the curtains. I am asking, one where we only know that it exists behind the curtains. I ask this because I want to make sure it is alright to say that we are just finding a delta as a function of epsilon when doing the proof – tryst with freedom May 15 '22 at 14:31
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    You wrote: "I want to make sure it is alright to say that we are just finding a delta as a function of epsilon when doing the proof". Well, of course that's okay. The underlying logic has no need for anything to be given by explicit formulas all the time. If your task is to prove theorems in abstract math, not to develop explicit error estimates for numerical analysis, then nobody cares about how $\delta$ depends on $\varepsilon$. So just move on with learning other stuff instead of getting hung up on this point. – KCd May 15 '22 at 21:58
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There are already a variety of good comments and answers from different points of view; I would like to present another perspective, and provide further details.


does there exist any epsilon delta proof, in which we can only show that the delta only exists (as is the minimum condition in the statement) rather than the explicit relation of it with epsilon?

No, as you guessed; (disregarding foundational issues that are bound to pop up eventually) any $\forall\epsilon,\exists\delta$ statement can be in principle turned into a more explicit relation between $\epsilon$ and $\delta$. Of course, in general how explicit the original setup is bounds from above how explicit this relation could be. Indeed, as discussed at Any function with a modulus of continuity proportional to any preassigned $\epsilon>0$ is Lipschitz-continuous, any $\forall\epsilon,\exists\delta$ is equivalent (disregarding foundations) to some $\exists\delta,\forall\epsilon$, at the cost of a jump in the type of $\delta$; more explicitly

$$[\forall\epsilon\in\mathbb{R}_{>0},\exists\delta\in\mathbb{R}_{>0}: P(\epsilon,\delta)\text{ is true }] \iff [\exists \delta\in F(\mathbb{R}_{>0};\mathbb{R}_{>0}),\forall \epsilon\in\mathbb{R}_{>0}: P(\epsilon,\delta(\epsilon)) \text{ is true}].$$

(In this context such a function $\delta$ is not unique; in contexts in which it is unique, $(\implies)$ with uniqueness is an implicit function theorem.)

More generally (in principle) one can convert (disregarding foundations) any

$$\forall \alpha\in A,\forall x_\alpha\in X_\alpha,\exists y\in Y: P(\{x_\alpha\}_{\alpha\in A},y)\text{ is true }$$

to a

$$\exists y\in F\left(\prod_{\alpha\in A}X_\alpha;Y\right),\forall \alpha\in A,\forall x_\alpha\in X_\alpha: P(\{x_\alpha\}_{\alpha\in A},y(\{x_\alpha\}_{\alpha\in A}))\text{ is true }.$$


why not just keep it in the statement of epsilon delta limit that "$\delta$ can be written as some function of $\epsilon $ , $\delta=f(\epsilon)$ with $f: \mathbb{R^+} \to \mathbb{R^+}$ "

This has to do with the scope and use of the statement to be written, which are ultimately based on personal preferences. For instance in the link above a uniform modulus of continuity for a function $f:X\to Y$ between metric spaces is defined as a function $\mathfrak{m}:\mathbb{R}_{>0}\to\mathbb{R}_{>0}$; however one could just as well define a modulus of continuity as a function $\mathfrak{m}:C^0_u(X;Y)\times \mathbb{R}_{>0}\to \mathbb{R}_{>0}$ such that for any $f\in C^0_u(X;Y)$, for any $\epsilon\in\mathbb{R}_{>0}$, and for any $x_1,x_2\in X$:

$$d_X(x_1,x_2)< \mathfrak{m}(f,\epsilon)\implies d_Y(f(x_1),f(x_2))<\epsilon.$$

(This perspective is useful e.g. when talking about equicontinuity.)

(Here $C^0_u$ is the collection of uniformly continuous functions.)

Another alternative is to consider $X$ and $Y$ as metrizable spaces. Define $\widehat{C^0_u}(X;Y)$ as the set of triples $(d_X,d_Y,f)$ such that $d_X$ is a distance function on $X$ compatible with the topology of $X$, $d_Y$ is a distance function on $Y$ compatible with the topology of $Y$, and $f:X\to Y$ is uniformly continuous w/r/t $(d_X,d_Y)$. Then one can define a modulus of continuity as a function $\mathfrak{m}:\widehat{C^0_u}(X;Y)\times \mathbb{R}_{>0}\to \mathbb{R}_{>0}$ such that for any $(d_X,d_Y,f)\in\widehat{C^0_u}(X;Y)$, for any $\epsilon\in\mathbb{R}_{>0}$, and for any $x_1,x_2\in X$:

$$d_X(x_1,x_2)< \mathfrak{m}(d_X,d_Y,f,\epsilon)\implies d_Y(f(x_1),f(x_2))<\epsilon.$$

(This perspective, although somewhat in jest, is useful when there are different distances involved. For instance if $f:X\to \mathbb{R}$ is $\theta$-Hölder with respect to a distance function $d:X\to X\to\mathbb{R}_{\geq0}$, then it is Lipschitz with respect to the distance function $d^\theta:(x_1,x_2)\mapsto d(x_1,x_2)^\theta$, i.e. $\mathfrak{m}(d,f,\epsilon)=\left(\dfrac{\epsilon}{C}\right)^{1/\theta} \implies \mathfrak{m}\left(C d^\theta,f,\epsilon\right)=\epsilon$; see the discussion at the above link; Semmes calls the operation $d\mapsto d^\theta$ the "snowflake functor"; see the reference at Exponents for Hölder functions on metric spaces . Note that pointwise moduli of continuity can be considered to be better suited for this "fixed topology, different distances" perspective.)

(Yet another alternative is to define a modulus of continuity as a function $\mathfrak{m}: \operatorname{Arr}(\operatorname{Met}_u)\times \mathbb{R}_{>0}\to \mathbb{R}_{>0}$, where $\operatorname{Arr}(\operatorname{Met}_u)$ is the collection (category) of all uniformly continuous functions from a metric space to another one...)


Of course it is rude to deconstruct/universalize/invariantize some concept if it's not absolutely necessary for what is to come. E.g. in hyperbolic dynamics certain objects are automatically continuous, almost never differentiable, but to further the theory one needs something more than continuity (see Why Do We Care About Hölder Continuity?). In general, areas like dynamics where matters like stability, approximation, perturbation and deformation naturally introduce more moving parts.

The discussion at Difference between soft analysis and hard analysis is also somewhat relevant.


Finally, to contextualize some of the comments above, fix a function $f:X\to Y$ between metric spaces, and define its "uniform $\epsilon-\delta$ bundle" as

$$E(f)=\{(\epsilon,\delta)\in\mathbb{R}_{>0}^2\,|\, \forall x_1,x_2\in X: d_X(x_1,x_2)<\delta\implies d_Y(f(x_1),f(x_2))<\epsilon\}$$

with (a priori partially defined; Notation for "function from a subset of $X$ into $Y$"? ) projection $\pi:E(f)\rightsquigarrow \mathbb{R}_{>0}$ onto the first coordinate. Then ($\pi$ is defined everywhere and) there is a global section of $\pi:E(f)\to \mathbb{R}_{>0}$ iff $f$ is uniformly continuous, and existence of such a global section is equivalent to the axiom of choice (see https://ncatlab.org/nlab/show/axiom+of+choice), if each fiber is nonempty. Note that with this framework, if there is a section $\delta: \mathbb{R}_{>0}\to E(f)$, any other function $\delta':\mathbb{R}_{>0}\to\mathbb{R}_{>0}$ whose graph is under the graph of $\delta$ will also be a section. Due to other inequalities needed in a given argument one may have a lower bound $\sigma: \mathbb{R}_{>0}\to\mathbb{R}_{>0}$ also, and it might be important, for what is to come, to establish not only a section $\delta$ that stays above $\sigma$, but also one that is measurable/continuous/differentiable etc.. It's clear that this framework too can be universalized in variety of ways.

(Although, as a disclaimer this framework is somewhat misleading; as one can often focus on countably many values for $\epsilon$ (and $\delta$) (in which case countable choice would be sufficient to produce a section), so that the geometry $\pi:E(f)\to \mathbb{R}_{>0}$ seems to suggest is quite flexible.)

Alp Uzman
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Answer in response to this comment from the OP on the excellent answer from @KCd .

I am asking, one where we only know that it exists behind the curtains. I ask this because I want to make sure it is alright to say that we are just finding a delta as a function of epsilon when doing the proof.

Whether or not that's OK depends on the context. For a graduate student or a professional mathematician, certainly. In an undergraduate course, perhaps. If the point of the question is to determine whether you can find an epsilon explicitly, then no. If the instructor or the text regularly says things like "this has been done before, no need to look behind the curtain" then yes.

When I ask my students for proofs, what I want to know is that they have convinced themselves for good reasons. How much handwaving is permitted depends on the level of the course and the student.

(When I was an undergraduate taking advanced calculus from Lars Ahlfors he informed us in the first class that he did not like proofs that reasoned from the hypotheses to some point, then back from the conclusion to the same point, and wrote "therefore" where the arguments met.)

Ethan Bolker
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