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When analysts say "$\epsilon$ (or whatever greek symbol) can be chosen arbitrary small", do they really just mean we can take $\epsilon = 0$ or $\epsilon \to 0$ later?

When I asked myself this question, I immediately began doubting my understanding of limits.

That is $\forall \epsilon>0 ,\exists \delta > 0 :|x-a|<\delta \implies |f(x) - L|<\epsilon$.

I was taught that a very long time ago, a limit is something you approach very closely, but not exactly equal to. So what logic rule am I breaking if i take $\epsilon = |f(x) - L|$?

Also, what is the advantage of proving two things are equal by saying they are epsilon close to each other? Isn't this really just make it harder than it needs it to be?

EDIT

I have also noticed that some other definitions that uses epsilon distance instead of just saying they are equal to each other.

$f \in R(\alpha)$ on $[a,b] \iff \forall \epsilon >0, \exists$ partition $P$ such that $$U(f,P,\alpha) - L(f,P,\alpha) < \epsilon$$

Now what is wrong with saying

$f \in R(\alpha)$ on $[a,b] \iff \exists$ partition $P$ such that $$U(f,P,\alpha) = L(f,P,\alpha) $$

Asaf Karagila
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Lemon
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    I can guarantee you that sometimes showing that $|x-y|<\varepsilon$ is far easier than sohwing directly that $x=y$, specially when $x$ and $y$ are defined as the "limit of something". – Josué Tonelli-Cueto Jun 23 '13 at 20:54
  • What it means is that for any given positive number $p$, we can choose $\epsilon$ such that $0\lt \epsilon \lt p$. – André Nicolas Jun 23 '13 at 21:21
  • @AndréNicolas, it is always bigger than $0$? – Lemon Jun 23 '13 at 22:23
  • Yes, it is always bigger than $0$. But in typical formulations we end up concluding that the quantity $|Q|$ is $\lt \epsilon$, and $|Q|$ can be $0$. – André Nicolas Jun 23 '13 at 23:00

3 Answers3

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More correctly, if $f(x)$ is a real-valued function defined in a real open interval around a real number $a$ (except possibly at $a$) then we say $\lim_{x\to a}f(x)=L$ if $$\forall\epsilon>0,\exists\delta>0:0<|x-a|<\delta\implies|f(x)-L|<\epsilon.$$

That $\epsilon$-$\delta$ language means that for any positive $\epsilon$, we can keep $f(x)$ within a distance $\epsilon$ of $L$ simply by making sure that $x$ stays within some distance $\delta>0$ of $a$ (except perhaps if $x=a$). Eliminating the $\epsilon$ and $\delta$ entirely, this means that we can keep $f(x)$ "as close as we like" to $L$, simply by keeping $x$ "close enough" to (but not equal to) $a$.

How close is "close enough"? Well, that will depend on several factors in general, and almost always on how close we'd like $f(x)$ to stay to $L$--if we want to keep $f(x)$ closer to $L$, we may need to keep $x$ closer to $a,$ too.

Note that $x$ is still a variable in this definition--unlike $L$ (and $a$), it is not fixed, so we can't simply take $\epsilon=|f(x)-L|.$

Now, if we fixed some $x_0\ne a$ and happened to know that $f(x_0)\ne L,$ then we could certainly take $\epsilon=|f(x_0)-L|.$ Then if $\lim_{x\to a}f(x)=L,$ we'd know that there was some $\delta>0$ such that $|f(x)-L|<\epsilon$ whenever $0<|x-a|<\delta$--that is, we could keep $f(x)$ closer to $L$ than $f(x_0)$ is, by making sure that $x$ stays within $\delta$ of (but not equal to) $a$. Note that such a $\delta$ will necessarily be less than $|x_0-a|,$ since if not, then $x_0$ is an $x$ that is within $\delta$ of (but not equal to) $a$, yet $|f(x_0)-L|=\epsilon$.

Let me sum up the discussion the previous paragraph: If there is some $x_0\ne a$ with $f(x_0)\ne L$, and we know that $\lim_{x\to a}f(x)=L,$ then we can keep $f(x)$ closer to $L$ than $f(x_0)$ is, by making sure that we keep $x$ closer to (but not equal to) $a$ than $x_0$ is by a sufficient amount.

Hopefully that summary doesn't surprise you or confuse you, but let me know if it does.

Does this help clear things up for you?

Cameron Buie
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  • Oh I forgot that $f(x)$ is variable. I guess this also kind of answer my edit. But could you spell it out and tell me what is the variable? I don't think the partition $P$ is since once i find that $P$, it is up to only $f$ again? – Lemon Jun 23 '13 at 22:41
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    I'm not entirely sure what you mean by $\alpha$, there, but it looks like you're discussing Riemann integrability over $[a,b]$. Consider the function $f(x)=x$ on the interval $[0,1].$ Take any partition $P,$ say into $n$ sub-intervals, with smallest interval length $\ell$. Then $U(f,P)-L(f,P)$ is the sum of the squares of the lengths of the subintervals (draw a picture), so $$U(f,P)-L(f,P)\ge n\ell^2>0.$$ – Cameron Buie Jun 23 '13 at 23:07
  • The definition was retrieved from Baby Rudin. But my question was (let's pretend you never posted your answer for now) why couldn't we just say $U(f,P) = L(f,P)$ – Lemon Jun 24 '13 at 00:11
  • Or with relevance to my question, what's wrong with choosing $\epsilon = U(f,P) - L(f,P)$? – Lemon Jun 24 '13 at 00:13
  • The reason you can't (or at least shouldn't) just say $U(f,P)=L(f,P)$ is that non-constant functions, like $f(x)=x$, would never satisfy that definition of integrability. – Andreas Blass Jun 24 '13 at 00:16
  • @sidht: Regarding the problem with requiring $U(f,P)=L(f,P),$ see Andreas's comment. The example I gave above shows that even super-nice functions would fail to be integrable under that altered definition. – Cameron Buie Jun 24 '13 at 00:20
  • @sidht: As for the reason we can't put $\epsilon=U(f,P)-L(f,P)$, it's because the partition will depend on the choice of $\epsilon>0,$ just as $\delta>0$ did in my answer. The idea of the partition $P$ is that we can choose subintervals that are "narrow enough" so that the difference between the upper and lower sums is "as small as we like." As in my answer, if $P_0$ is a partition such that $U(f,P_0)>L(f,P_0),$ then we can put $\epsilon_0=U(f,P_0)-L(f,P_0),$ and if $f$ is Riemann integrable, there is a partition $P$ such that $U(f,P)-L(f,P)<\epsilon_0$. ... – Cameron Buie Jun 24 '13 at 00:22
  • ... In other words, if the upper sum and lower sum of a Riemann integrable function aren't the same for a given partition, then we can find a partition that makes the upper sum and lower sum closer. – Cameron Buie Jun 24 '13 at 00:24
  • @anyone, regarding the $f(x) = x$ example, then how does $U(f,P) - L(f,P) < \epsilon$ fixes the problem? Aren't we going to send $\epsilon \to 0$ anyways? Because we have another problem now. – Lemon Jun 24 '13 at 00:33
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    We are not sending $\epsilon$ anywhere. Rather, we are choosing an arbitrary positive $\epsilon$ which will be fixed, then showing that there is some partition $P$ (which depends on the $\epsilon$ we chose), so that the upper and lower sums differ by less than $\epsilon.$ To do this (in my example above), find $n$ sufficiently large so that $\frac1n<\epsilon,$ and let $P$ be the partition of $[0,1]$ into $n$ equal subintervals of length $\ell=\frac1n.$ Then $$U(f,P)-L(f,P)=n\ell^2=\frac1n<\epsilon.$$ – Cameron Buie Jun 24 '13 at 00:39
  • But isn't the goal still make $\epsilon$ small? – Lemon Jun 24 '13 at 01:21
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    @sidht: Not really. It is to show that we can choose $\epsilon>0$ to be as small as we like, and still find a partition $P$ to make $U(f,P)$ and $L(f,P)$ differ by less than $\epsilon$. We get our flexibility in $\epsilon$'s size by choosing it in the first step. – Cameron Buie Jun 24 '13 at 01:27
  • Right and isn't that what the analysts mean when they say as small as we like (and later send $\epsilon \to 0$)? – Lemon Jun 24 '13 at 01:54
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    @sidht: Analysts who say "send $\epsilon\to 0$" are being sloppy, usually because they are in a situation where they know they can afford to be. This tends to lead to confusion in people who are newer to the concepts, though, so it's a terrible thing to say from a teaching standpoint. – Cameron Buie Jun 24 '13 at 01:56
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They certainly do not mean you can take $\epsilon=0$. Usually, the statement

$\epsilon$ can be chosen arbitrarily small

means that whatever property the statement is referring true holds for any $\epsilon>0$. Often this fact is used to take some limit as $\epsilon\to 0^+$ (called the limit from the right as $\epsilon$ goes to $0$).

Alex Becker
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  • What is my mistake in my post then? I know there is something wrong when I chose $\epsilon = |f(x) - L|$ – Lemon Jun 23 '13 at 21:11
  • @sidht I don't know what you're doing. Your definition of limit is correct, but all that is is define what $\lim\limits_{x\to a}f(x)=L$ means. You don't set $\epsilon$ equal to anything in the definition. – Alex Becker Jun 23 '13 at 21:14
  • Why not? I know the distance between $f(x)$ and $L$ are epsilon close, and since this is true for each $\epsilon >0$. I chose that $\epsilon$. – Lemon Jun 23 '13 at 21:56
  • @sidht I'm very confused. It's a definition. You don't "do" anything. – Alex Becker Jun 23 '13 at 21:58
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    @sidht when you say "the distance between $f(x)$ and $L$" you already make the crucial mistake. $L$ is a number and $f(x)$ is a function, so there is no such thing as the distance between one and the other. The definition of limits does not claim that the distance between a function and a number is 0. Intuitively what it says is that the value $f(x)$ can be made arbitrarily close to $L$ by making $x$ arbitrarily close to $a$, where $a$ is where the limit is computed. Formally, the definition of limit is the definition of limit. – Ittay Weiss Jun 23 '13 at 22:39
  • @IttayWeiss, right what I meant was the distance between $f$ and $L$ can be made arbitrarily close when $x$ is close to $a$ – Lemon Jun 23 '13 at 22:46
  • @sidht Not when $x$ is close to $a$, but by setting $x$ sufficiently close to $a$ – Alex Becker Jun 23 '13 at 22:48
  • @AlexBecker, are you trying to correct me by saying it is very close instead of close because that's usually what I mean when I say close. – Lemon Jun 23 '13 at 22:52
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    @sidht you totally miss the point. There is no distance between $f$ and $L$. The quantity $|f(x)-L|$ can be made arbitrarily small by setting the quantity $|x-a|$ to be sufficiently small. It's all about approximations. – Ittay Weiss Jun 23 '13 at 23:09
  • Am I allowed to replace the word quantity with distance? – Lemon Jun 23 '13 at 23:12
  • @sidht The distance between $f(x)$ and $L$ is the quantity $|f(x)-L|$, and similarly the distance between $x$ and $a$ is the quantity $|x-a|$. – Alex Becker Jun 23 '13 at 23:47
  • @AlexBecker, yeah that's what meant. Isn't that what I was saying when I said "the distance between $f(x)$ and $L$"? – Lemon Jun 24 '13 at 00:02
  • @sidht We're getting confused between when you are referring to the value of $f$ at some fixed $x$ and when you are referring to the function $f$. – Alex Becker Jun 24 '13 at 00:12
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    Oh you guys were thinking that I was saying the distance between a number and a fluctuating (maybe not the best adj here) function? – Lemon Jun 24 '13 at 00:34
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This is really more of an extended comment, rather than an answer.

From the point of view of pure logic, the phrase "$\epsilon > 0$ can be chosen arbitrarily small" simply means "for all $\epsilon > 0$." That's it.

So why include the "arbitrarily small" part?

The phrase "arbitrarily small" is simply telling the reader how s/he should think about the property involving $\epsilon$. Namely, it tells the reader that it will be relatively easy to satisfy the property with large values of $\epsilon$, but harder to satisfy it with smaller values of $\epsilon$. And yet, despite this difficulty when small values of $\epsilon$ are taken, the statement is still true!

Example: Suppose I tell you that for every $\epsilon > 0$, there exists a real number $\delta$ satisfying $0 < \delta^3 < \epsilon$. You're not sure whether to believe me, so you try examples:

  • If $\epsilon = 1$, then there are lots of numbers $\delta$ that make the statement true. (e.g. every $0 < \delta < 1$ works)

  • If $\epsilon = 100$, then there are even more values of $\delta$ that make the statement true. (e.g. every $0 < \delta < 4$ works)

In fact, if you've proven that my statement is true for a given $\epsilon$, then it's also true for all greater $\epsilon' > \epsilon$. Like, if it's true for $\epsilon = 1$, then it's true for $\epsilon = 2.5, 17, 10000$, etc.

But I said that it's true for all $\epsilon > 0$. That means that we have to look at small values of $\epsilon$, too. And in these cases, values of $\delta$ that work are less plentiful.

In such a situation, I would say that my statement is true for all $\epsilon >0$, no matter how small, to emphasize this point.

Note, however, that although my statement is true for all $\epsilon > 0$, no matter how small, it is decidedly false for $\epsilon = 0$.

Community
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Jesse Madnick
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  • Okay I have to be honest, out of all the answers it seems like you understand me the most and I am not easiest student to understand. – Lemon Jun 26 '13 at 04:09
  • SO basically you are saying $\forall \epsilon > 0$ is in some ways, a "without any loss of generality" or some sort? – Lemon Jun 26 '13 at 04:10
  • If I say that "$\forall \epsilon > 0$, the statement $P_\epsilon$ is true," then I mean just that: The statements $P_1$, $P_{15}$, $P_{0.17}$, etc are all true. The phrase "without loss of generality" (WLOG) means something different. WLOG is a phrase used during a proof to indicate that "If we prove a certain simple case, then the general case will follow." – Jesse Madnick Jun 26 '13 at 08:24
  • Example: If you're trying to prove the example in my above answer, then -- since we proved the case of $\epsilon = 1$, and we also know that the statement will therefore be true for all $\epsilon > 1$ -- we can assume WLOG that $\epsilon < 1$. It then remains to prove that the statement is true for all $0 < \epsilon < 1$. Of course, in this case that doesn't really help us, but in some situations this sort of simplification can really help. In my opinion, though, I think beginners should avoid using the phrase "WLOG" in their own proofs. – Jesse Madnick Jun 26 '13 at 08:27
  • I put "WLOG" in quotations on purpose, but I am trying to connect an analogy. I am saying that by showing it is true for $\forall \epsilon$, no matter how big or small, you have shown for it is true for each $\epsilon$. I know this is a very stupid sentence. – Lemon Jun 27 '13 at 05:55
  • Again, the words "small" and "large" are purely psychological; they tell you how to intuit what you're doing. You can do analysis just fine without ever using either term. But yes... if you've proven that a statement holds $\forall \epsilon > 0$, then, yes, it certainly is true for EVERY $\epsilon > 0$. – Jesse Madnick Jun 27 '13 at 06:29