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Consider the "formal definition" here https://en.wikipedia.org/wiki/Limit_of_a_sequence. I checked some references and this is often precisely the definition in all words and terms used in this article. I claim that this definition is not precise. And I am sure I am wrong. That is why I would like to ask about it. Consider the "illustration" part. This figure makes the concept very clear. It shows that as $\varepsilon$ decreases $N$ should increase. So this means that (i) $N$ must be a function of $\varepsilon$, and (ii) $N$ must be a decreasing function of $\varepsilon$. Why are these not made part of the formal definition by at least writing $N(\varepsilon)$ instead of $N$ and perhaps adding a sentence that $N$ must be a decreasing function of $\varepsilon$? If these are not made explicit, why is it clear from the formal definition that $x_n$ actually converges to $x$ and $n$ increases?

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    $\begingroup$ $N$ is never unique. $\endgroup$
    – Randall
    Oct 15 '18 at 12:37
  • $\begingroup$ Also consider constant sequences. $\endgroup$
    – Randall
    Oct 15 '18 at 12:42
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    $\begingroup$ Perhaps you are thinking of $N$ as the minimal integer such that $|x_n - x|< \epsilon$ for $n \ge N$. But that is not a practical definition. $\endgroup$
    – Martin R
    Oct 15 '18 at 12:44
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    $\begingroup$ It can be clarifying to write something like: "... if for every $\epsilon>0$ there is an integer $N_{\epsilon}$ such that..." This to emphasize that this integer depends on $\epsilon$ up to a certain level. It is not determined by $\epsilon$ so we cannot speak of a function. Existence is enough. $\endgroup$
    – drhab
    Oct 15 '18 at 12:56
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    $\begingroup$ "Up to a certain level" because every $\epsilon>0$ determines an integer $K_{\epsilon}$ such that every $n\geq K_{\epsilon}$ is a candidate for $N_{\epsilon}$. Here $K_{\epsilon}$ is the smallest integer that "works" for $\epsilon$ and indeed $\epsilon\mapsto K_{\epsilon}$ is a function. So there is a restriction for $N_\epsilon$: it cannot be smaller than $K_{\epsilon}$ (determination up to a certain level). But no determination: it can be any integer that is not smaller than $K_{\epsilon}$. $\endgroup$
    – drhab
    Oct 15 '18 at 13:12
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You are right that $N$ depends on $\epsilon$ - and missing this fact is the cause of a lot of confusion among people learning analysis for the first time. Sometimes it is made explicit: "for all $\epsilon > 0$ there exists an $N$ (dependent on $\epsilon$) such that...". But even when phrased as "for all $\epsilon > 0$ there exists an $N$ such that...", $N$ still depends on $\epsilon$: that's how quantifiers ("for all", "there exists") work, even if it's not made explicit.

One thing that should be pointed out is that $N$ is not unique. Indeed if, for a given $\epsilon$, $N = M$ works[1], then $N = M'$ works for any $M' \geq M$. We could define things such that $N$ was the minimum that worked, but this would make both the definition and proofs involving the definition more convoluted, with very little gain in return.

With this in mind, it's somewhat unhelpful to think of $N$ as a function of $\epsilon$, because "function" suggests there's a unique value of $N$ for each $\epsilon$, which isn't true.

On the topic of $N$ increasing as $\epsilon$ decreases: this is not true in the most pedantic sense: if $x_i = 0$ for all $i$, then we could take $N = 10$ when $\epsilon > 1$ and $N = 1$ when $\epsilon \leq 1$, the definition is satisfied but $N$ is not increasing as $\epsilon$ decreases.

However, in a less pedantic sense, it is true that if $N$ works for $\epsilon$ and $\epsilon' > \epsilon$ then $N$ works for $\epsilon'$ too. But this is a consequence of the definition, and so doesn't need to be included in the definition.

To answer your third question, "why is it clear from the formal definition that $x_n$ actually converges to $x$ as $n$ increases?", there's a temptation to ask "what do you mean by 'actually converges'?" Perhaps a helpful observation is that the definition is equivalent to saying: for any $\epsilon > 0$, all but finitely many of the terms of the sequence are within $\epsilon$ of $x$. We don't care how many that "finitely many" is, provided there are only finitely many of them.

[1]: I hope it's clear what I mean by "$N$ works for $\epsilon$" here and later, I just don't want to have to keep writing out the second half of the definition

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    $\begingroup$ Of course, one could phrase the definition as "There exists a function $N(\epsilon)$ such that for all $\epsilon$, if $n>N(\epsilon)$ then $|x_n-x|<\epsilon$. $\endgroup$ Oct 15 '18 at 15:12
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    $\begingroup$ @Snoopy While the $n$ that we choose need not be a "function" of $\epsilon$, a function of $\epsilon$ must exist. $\endgroup$ Oct 15 '18 at 17:20
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    $\begingroup$ @Snoopy if an $N$ exists for each $\epsilon$, then you can pick one such $N$ for each $\epsilon$ and call that $N(\epsilon)$, and then write out all the definitions in terms of that function $N(\epsilon)$. But (a) you don't need to and (b) because that function isn't uniquely defined, it results in cleaner proofs not to do so. $\endgroup$ Oct 16 '18 at 8:42
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    $\begingroup$ @Snoopy I don't think any of that is wrong. The issue with the notation $N_\epsilon$ is that $N$ also depends on the sequence. For example, let's say we're trying to prove: if $a_n \to a$ and $b_n \to b$ then $(a_n + b_n) \to a + b$. We pick $\epsilon > 0$. Then there exists $N'$ such that if $n > N'$, $|a_n - a| < \frac\epsilon 2$ and $N''$ such that if $n > N''$, $|b_n - b| < \frac\epsilon 2$. Then set $N = \max N', N''$, then for $n > N$, $|(a_n + b_n) - (a + b) \leq |a_n - a| + |b_n - b| < \frac \epsilon 2 + \frac \epsilon 2 = \epsilon$. $\endgroup$ Oct 16 '18 at 10:43
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    $\begingroup$ (continued) We have to make it clear that $N$, $N'$, $N''$ are all different anyway, because they're for different sequences, adding the subscripts $N_\epsilon$, $N'_{\frac \epsilon 2}$, $N''_{\frac \epsilon 2}$, adds noise for no gain. $\endgroup$ Oct 16 '18 at 10:43
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$N$ need not be a function of $\epsilon$. For a given $\epsilon$, many $N$ are possible (in particular all values larger than $N$ also work).

Enforcing $N$ to be some function a $\epsilon$ wouldn't be a good idea, because it would impose a constructive solution, i.e. require you to exhibit the particular value of $N(\epsilon)$. Though in reality, it suffices to prove that a suitable value exists, whatever it is.

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  • $\begingroup$ Also, having "$N$ must be a decreasing function of $\varepsilon$" as part of the definition means this will have to be verified in proofs, or reference will have to be made to a(n unnecessary) theorem that says without this assumption the result still holds. $\endgroup$ Oct 15 '18 at 12:49
  • $\begingroup$ @Yves (1) Why is it true that if I decrease $\varepsilon$ I necessarily get a subset of the possible $N$? (2) This claim also makes it clear that $\varepsilon$ and $N$ should have a link. Right? $\endgroup$
    – Snoopy
    Oct 15 '18 at 12:56
  • $\begingroup$ To the upvoters: I thank you, but I have completely rewritten since. Feel free to unupvote if you don't like the new version. $\endgroup$
    – user65203
    Oct 15 '18 at 13:23
  • $\begingroup$ I think @DaveL.Renfro first comment is the best point here. $\endgroup$
    – Randall
    Oct 15 '18 at 15:00
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The wikipedia article states that we call $x$ the limit of the (real) sequence $(x_n)$ if the following condition holds:

$∀ ε > 0 ( ∃ N ∈ \mathbb N ( ∀ n ∈ \mathbb N ( n ≥ N ⇒ | x_n − x | < ε ) ) )$.

Which is, contrary to your claim, is precise and complete. You are not interpreting the basic logical syntax correctly, which is why you do not understand that the order of quantifiers completely determines the correct interpretation. Here in particular, the "$∃ N ∈ \mathbb N (\cdots)$" is under the "$∀ ε > 0$", so the $N$ that is claimed to exist is not necessarily the same for different $ε$.

Also, there is no logical or pedagogical need to think of $N$ as a function of $ε$. But if you wish, you can do so via Skolemization. However, I recommend you wait until you learn first-order logic before you go into that.

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  • $\begingroup$ Yes I struggle to read the stated expression. It would be helpful if someone could write it out in words. On the other hand, I already acknowledged that the claim $N$ is a function of $\varepsilon$ is not correct. $\endgroup$
    – Snoopy
    Oct 15 '18 at 18:15
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    $\begingroup$ @Snoopy: That's precisely the problem. You have to learn this basic logic, because it underlies all of mathematics whether or not it is made explicit. A common mistake is to think that logical syntax can be read in a glance; it can't because it is a 100% precise but highly compressed statement. A statement "∀x∈S ( P(x) )" says that "P(x)" holds for every x in S. "P(x)" in turn could involve a quantifier, but the point is that is is asserted separately for each x in S. Rewriting the whole sentence in natural language will not help you much. Wikipedia did just that, and you didn't get it. $\endgroup$
    – user21820
    Oct 15 '18 at 18:24
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    $\begingroup$ @Snoopy: I'll note that Wikipedia's English version is already as precise as you can get in English, and it conforms to perfectly grammatical standard English. An analogous example is: "For each citizen X of the United States, there exists a state S, such that for majority of X's life, X has lived in S.". $\endgroup$
    – user21820
    Oct 15 '18 at 18:29

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