This is a question regarding Epsilon-Delta proofs, in this example in Single-Variable Calculus, but hopefully the crux of what I'm asking here, will be general enough.


I'm putting this loose introduction here to illustrate the point I'm trying to make about $\epsilon - \delta$ proofs below.

Given a function $f$, it's output $f(x)$ is dependent on its input $x$. Therefore given an arbitrary output $f(x)$ (which we don't know if it exists in the functions Co-Domain) we need to prove that there is a corresponding $x$ that produces it, otherwise the function $f$ would be undefined at that point $x$

However given an $x$ that is in the domain of $f$, we can ALWAYS prove that there exists a $f(x)$, as the Co-Domain of a function is entirely dependent on its Domain.

Onto the $\epsilon - \delta$ Part

The formal definition of a limit (the $\epsilon-\delta$ definition) is :

$$ \lim_{x \to a} f(x) = L \Leftrightarrow \forall \epsilon >0 (\exists \delta>0 \ni (\forall x \in D(0 < |x-a| < \delta \implies |f(x)-L| < \epsilon)))$$

Essentially what this is saying that as we restrict $x$ to being within $\delta$ units of $a$, then as a result of that restriction $f(x)$ becomes restricted to being withing $\epsilon$ units of $L$

You can see this visually below.

enter image description here

Now naturally I would assume that because of this, it would mean that $\epsilon$ is dependent on $\delta$, i.e. as we restrict $x$ to being within $\delta$ units of $a$, then as a result of that restriction $f(x)$ becomes restricted to being withing $\epsilon$ units of $L$. But my assumption is wrong.

A quote from Wikipedia :

Therefore δ depends on ε. The limit statement means that no matter how small ε is made, δ can be made small enough.

Even from the formal definition I referenced it states : $\forall \epsilon > 0, \exists\delta > 0$, implying once again that $\delta$ is dependent on $\epsilon$.

Why is that so? Why does the Epsilon-Delta definition of a limit, treat values of $\epsilon$ which can be elements of a functions Co-Domain, as being independent, and values of $\delta$ which can be elements of a functions Domain as being dependent on $\epsilon$?

Or am I wrong, and does $\epsilon$ and $\delta$ not represent elements of a functions co-domain or domain (i.e. they have nothing to do with the functions domain and co-domain), but rather concepts of distances? With $\epsilon$ being the "error distance", and $\delta$ being the "dependent distance".

Another quote from Wikipedia :

In these terms, the error ($\epsilon$) in the measurement of the value at the limit can be made as small as desired by reducing the distance ($\delta$) to the limit point.

But in this quote it seems that $\epsilon$ is dependent on $\delta$ as it says the error ($\epsilon$) can be made as small as possible, by reducing the distance ($\delta$), implying $\epsilon$ is dependent on $\delta$, so which is it?

To Summarize

I know this is a long post, but I've written all of this out to clarify my thought processes so you can see where I'm coming from.

In a nutshell I am asking two separate questions?

  1. Are $\epsilon$ and $\delta$ meant to be thought of purely as distances? (which I'm sure they are, otherwise contradictions pop up)

  2. Is $\epsilon$ dependent on $\delta$, or is $\delta$ dependent on $\epsilon$? i.e. is the "error distance" $\epsilon$ dependent on the "restricting distance" $\delta$ or is it the other way around?

If you have found any gaps, in my knowledge or understanding, please inform me, as an undergraduate student majoring in Pure Mathematics, I'm always looking to improve my understanding. Secondly if a deeper understanding on mathematically rigorous definition of limits, can be found by concepts in Real Analysis, I'd love to hear about them.


Indeed both $\epsilon$ and $\delta$ are distances.

Maurice Fréchet generalized this idea further by defining Metric spaces. For example on $\mathbb R$ the usual metric is called the Euclidean metric where the distance of $x, y \in \mathbb R$ is $|x-y|$.

So $|f(x) -L| < \epsilon$ means the distance of $f(x)$ from $L$ is less than $\epsilon$.

You are correct, by definition $\delta$ depends on $\epsilon$.

A better wording for your second quote in my opinion would be the following:

One can make the error arbitrarily small by choosing a small enough $\delta$.

See also this question.

Hope this helps.

  • $\begingroup$ If $\delta$ depends on $\epsilon$ is it fine to think of delta as a function of epsilon i.e. $\delta(\epsilon)$? $\endgroup$ Apr 17 '16 at 17:54
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    $\begingroup$ If for a given $\epsilon$ a $\delta$ is "good", then any $0 < \delta' <\delta$ is also "good". So I'm not sure it'd be a well-defined function, but you can certainly think of it as one - keeping in mind that there are a lot of other $\delta$s for that $\epsilon$. $\endgroup$ Apr 17 '16 at 18:02
  • $\begingroup$ @Andrew, Is it possible to prove $0 < \delta' <\delta$ is also ''good'', by definition ? I don't think so. Hence, in my opinion the given definition is incomplete. Or am I wrong ? $\endgroup$ Jun 29 '16 at 8:15
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    $\begingroup$ @HardeyPandya I say a $\delta$ is "good" if and only if $|x - a| < \delta \Rightarrow |f(x) - L| < \epsilon$ is true. Assume $\delta$ is "good". Pick any $0<\delta'<\delta$. Then $|x - a| < \delta' \Rightarrow |x - a| < \delta$, since $\delta' < \delta$. So we have that $|x - a| < \delta' \Rightarrow |x - a| < \delta \Rightarrow |f(x) - L| < \epsilon$ is true. Finally by the transitivity of implication, $|x - a| < \delta' \Rightarrow |f(x) - L| < \epsilon$ thus $\delta'$ is also "good". $\endgroup$ Jul 3 '16 at 12:17
  • $\begingroup$ Perfect. @Andrew. $\endgroup$ Jul 10 '16 at 10:32

(1).In set theory a function $f$ from $A$ to $B$ (written $f:A\to B$ ) is defined as any set $f\subset A\times B$ such that $$\forall x\in A\;\exists! y\in B\;(\;(x,y)\in F)\}.$$ (In case you don't know, "$\exists!y$" means "there exists exactly one $y$".) And we write $y=f(x)$ to mean $(x,y)\in f.$ That is, the function $f$ is equal to "its graph". The set $A$ is the domain of $f. $ The shortcoming of this def'n that if $f$ is a function from $A$ to $B$ it is also a function from $A$ to $C$ whenever $C\supset B.$

The co-domain of $f$ is defined as $$\{y:\exists x\;(\;(x,y)\in f\;)\}.$$ It is illogical to refer to an "output" $y$ as $f(x)$ unless $(x,y)\in f.$ The co-domain can also be written as $\{f(x):x\in A\}.$ Set-theorists also use the notation $f$''$D$ (read $f$ double-tick $D$.): $f$''$D =\{f(x):x\in D\cap$ dom$ (f)\}.$ This is mostly used when $D\subset$ dom $(f),$ when $f$"$D=\{f(x):x\in D\}.$ So the co-domain of $f$ is $f$''dom $(f).$ This notation avoids the ambiguity of the notation $f(D)$ or $f D,$ when a set $D\subset$ dom $(f)$ is also a member of dom $(f).$

(2). epsilon/delta. Since we say "For each $\epsilon$ there exists a $\delta$ such that $S(\delta,\epsilon )$" the range of allowable values of $\delta$ is dependent on $\epsilon.$ It is also often desirable, or needed, to know what values of $\epsilon$ (if any), for a given $\delta,$ will satisfy $S(\delta,\epsilon ).$


The way you have put the definition of limit is the key to your problems. Using logical symbols makes things precise but it comes at a huge cost in terms of making things very difficult to understand. The same thing could be written in language of communication (English here) without any loss in precision and would have led to far less confusion. The only cost involved is that of typing/writing/adding more pages in book.

The fact that $f(x) \to L$ as $x \to a$ means that the values of $f$ can be made to lie as close to $L$ as we want (this leads to arbitrary $\epsilon > 0$ because our wants are always arbitrary) by choosing values of $x$ sufficiently close to $a$ (this leads to $\delta > 0$ and it is a means to satisfy our wants). Because the primary goal is to make values of $f$ close to $L$, it is clear that the means ($\delta$) to achieve the goal ($\epsilon$) will always be dependent on the goal. The limit definition emphasizes the fact that however tough the goal we set (meaning take smaller and smaller values of $\epsilon$), it will possible to find the means to achieve the goal.

Another fallacy which you have is about the co-domain part. We don't mean to say that every number close to $L$ is a value of $f$, but rather that every value of $f$ can be made close to $L$ by choosing values of $x$ near $a$.

The role played by $\epsilon, \delta$ is quantifying the closeness of things being talked about (closeness of $f(x)$ to $L$ and closeness of $x$ to $a$) and in this sense they actually control the distances $|f(x) - L|$ and $|x - a|$. And I believe that using greek symbols makes them more mysterious.

Another source of confusion comes from the implication $$0 < |x - a| < \delta \Rightarrow |f(x) - L| < \epsilon$$ which seems to suggest that $\delta$ is something primary and by implication it leads to $\epsilon$. However this is exactly the opposite of what the definition says. The definition requires to choose an $\epsilon$ first and then asks us to find a $\delta$ which will make this implication true. The definition guarantees that it will be always be possible to find the $\delta$ (given $\epsilon$) which makes this implication true.

See another answer (which is almost the same as here) of mine for a similar question.

Update: It appears that my answer linked above is already referred to by OP in his answer. I did not see this earlier and typed an almost similar answer here. Will keep it posted here so as not to waste my efforts of typing.

  • $\begingroup$ I just can't understand how the expression $(\forall \epsilon > 0) (\exists \delta >0)$ tells us the method of finding delta. $\endgroup$ Jun 29 '16 at 8:24
  • $\begingroup$ @HardeyPandya: Yes the definition does not tell how to find delta. Purpose of the definition is to define and not to describe strategy for effective usage. Effective use of definition requires some experience with actual examples. There is no fixed strategy to find the $\delta$ and you need to learn it by solving more and more problems. $\endgroup$
    – Paramanand Singh
    Jun 29 '16 at 9:24
  • $\begingroup$ That's why I think, this definition is incomplete. Or I am wrong ? $\endgroup$ Jun 29 '16 at 9:27
  • $\begingroup$ @HardeyPandya: No a definition is not necessarily required to include a strategy for effectively using it. Thus a definition of GCD of two numbers does not mention the Euclidean algorithm to find the GCD. $\endgroup$
    – Paramanand Singh
    Jun 29 '16 at 9:31
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    $\begingroup$ @HardeyPandya: Because we are defining limit we are only supposed to give the meaning of limit. $\delta$ is just a positive real number and there is no need to mention anything further about it. I am not able to understand what more about $\delta$ you want to include in the definition. $\endgroup$
    – Paramanand Singh
    Jun 29 '16 at 9:36

Thank you all for submitting answers however the most intuitive answer I found was located here in the list of answers to a very similar question asked by another user : Why can't epsilon depend on delta instead?

Specifically, this was the answer that I found the most intuitive : https://math.stackexchange.com/a/1324644/266135


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