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I'm trying to do the following exercise for my Real Analysis class:

Let $p$ be a given natural number. Give an example of a sequence $\left(x_{n}\right)$ that is not a Cauchy sequence, but that satisfies $\lim \left|x_{n+p}-x_{n}\right|=0$

However, I am in constant doubt in regard to the "let" word in math texts. Can I choose, say $p = 1$, or when one says "let" I am supposed to stick with $p \in \mathbb{N}$ and nothing more? What kind of "control" do I have over $p$?

For instance, when one says "Given $\epsilon \gt 0$", I usually see things like: Let $\epsilon = \frac{\epsilon}{2}$ so one can finish a certain argument.

Can someone help me? I need to solve this doubt once and for all, it bothers me very often.

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    $\begingroup$ You have no control over $p$. The author has introduced $p$ and told you that it is a natural number, and has given you no additional information about $p$. $\endgroup$ – littleO Oct 30 at 4:54
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    $\begingroup$ Does Bartle actually say "Let $\epsilon=\epsilon /2$" after stating that $\epsilon>0$? $\endgroup$ – DanielWainfleet Oct 30 at 5:04
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    $\begingroup$ @DanielWainfleet: If Bartle does that, then Bartle is wrong. That's just nonsense. Please, Lucas, clarify whether Bartle really does that. $\endgroup$ – user21820 Oct 30 at 16:13
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    $\begingroup$ And I strongly suggest you read carefully through this post about "let". $\endgroup$ – user21820 Oct 30 at 16:14
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    $\begingroup$ Bartle does not do that @user21820 I got confused because sometimes $\frac{\epsilon}{2}$ just pops up. However, that does not happen like i said. Edited my question in that matter $\endgroup$ – Lucas Oct 30 at 16:44
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To attempt an answer to your question, any time you read an instance of "Let $x$ be an element of $S$," you are being asked to consider an element $x$ whose only salient property as far as you are concerned is that it is a member of $S$. Until further assumptions are made about the specific nature of $x$, you should only use properties of $x$ that it has as an element of $S$ (and any supersets that $S$ is contained in, depending on the context). Some examples:

  1. Let $x$ be a real number. (Or "Let $x$ be an element of $\mathbb R$" to match the language I used above.)
  • This means to only consider the properties that $x$ has as a real number, or possibly also as a complex number, for example depending on the situation.
  1. Let $n$ be a given natural number.
  • This means that for our discussion, $n$ is a natural number, but that is all we know about it. Depending on what we talk about next, we may or may not also use other properties of $n$, like that it is a rational number, or that it is a complex number, for instance.
  1. Let $A$ be a subset of $\mathbb R$.
  • This is similar to the first example. All we want to do at this point is consider the properties of $A$ we know it has as a subset of $\mathbb R$. It may be the collection of natural numbers, it may be the empty set, it may be $\{\pi, \pi^2\}$, and so on.

Part of the point of making statements like this is that if we then go on to prove something involving this arbitrary element $x$, then we can substitute the value of $x$ with any specific element of $S$ and obtain a "new" theorem. For your example, if you can prove the statement you are given assuming nothing other than $p$ is a natural number, then you know, for example that there is a sequence $\{x_n\}$ in $\mathbb R$ that is not Cauchy, with the property that $$ \lim_{n\to\infty}(x_{n+314159}-x_n) = 0, $$ which looks pretty impressive, but is really nothing other than a special case where $p = 314159$ of the theorem stated in terms of the otherwise arbitrary $p$. We get to know that this "$314159$ theorem" is true for free if the only property of $p$ we use in our proof is that $p$ is a natural number since $314159$ is also a natural number.


Edit: "The $\epsilon$ issue." In analysis, we often want to prove theorems that have the form "For all $\epsilon > 0$, $P(\epsilon)$ is true." Where $P(\epsilon)$ is a statement involving $\epsilon$. For example,

  1. $P(\epsilon) = $ there exists $\delta > 0$ so that $|x^2-100| < \epsilon$ if $|x-10| < \delta$.

  2. $P(\epsilon) = $ there exists $N\in\mathbb N$ so that for all $n,m\ge N$, $|x_n-x_m| < \epsilon$.

When you think about these theorems in this way, I think it becomes a little bit clearer why in order to establish the truth of $P(\epsilon)$ for all $\epsilon > 0$, it suffices to establish the truth of $P(\epsilon/2)$ for all $\epsilon > 0$. Essentially, $\epsilon$ is a place-holder for "all the positive real numbers," and for that purpose, if $C$ is any constant like $2$ or $\frac12$, then $C\epsilon$ is also a place-holder for all the positive real numbers if $\epsilon$ is a place-holder for all positive real numbers. So it suffices to show that for all $\epsilon > 0$, $P(C\epsilon)$ is true. (It is critical however that $C$ does not depend on $\epsilon$! For instance, if we tried to let $C$ depend on $\epsilon$, then we could "cheat" and prove $P(\frac1\epsilon \epsilon)$ is true for all $\epsilon > 0$, but this only shows that $P(1)$ is true...)

Edit to the edit: I wrote in parentheses at the end of my last edit that $C$ is not allowed to depend on $\epsilon$, but this is not quite correct. So long as $C\epsilon$ is a place-holder for all the positive real numbers, $C$ can itself be a function of $\epsilon$. Here are two examples and one non-example of when $C\epsilon$ is a place-holder for all the positive real numbers when $\epsilon$ is itself a place-holder for all the positive real numbers (you can try to think of more examples for fun!):

  • $C = (1+\epsilon^2)$. Then $C\epsilon = \epsilon + \epsilon^3$ is still a place-holder for all the positive real numbers.

  • $C = \frac{1}{\sqrt{\epsilon}}$. Then $C\epsilon = \sqrt{\epsilon}$ is still a place-holder for all the positive real numbers.

  • $C = \frac{1}{\epsilon}$. Then $C\epsilon = 1$ is not a place-holder for all the positive real numbers. This was the kind of "malicious" dependence on $\epsilon$ that I had in mind when I said that $C$ should not depend on $\epsilon$ in the parenthetical sentence at the end of my first edit.

Sometimes it can be convenient to allow $C$ to depend on $\epsilon$. I know of at least one example where the bound I had by the end of the proof was $\epsilon + \epsilon^3$, which motivates my first example I wrote here where $C = 1 + \epsilon^2$.

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  • $\begingroup$ Interesting and super clear. What about the $\epsilon$ thing i wrote about? It really makes me uncomfortable. I tend to think that the claim should be proved for $\epsilon > 0$ and that's it. I don't get why one has the right to choose a certain $\epsilon$ $\endgroup$ – Lucas Oct 30 at 4:41
  • $\begingroup$ @Lucas: I updated my answer a little bit to furnish more examples and also address "the epsilon issue." Hopefully it helps! $\endgroup$ – Alex Ortiz Oct 30 at 4:54
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    $\begingroup$ Well, this comment box right here tells me i should not use it only to say "Thanks". However, there is just no way i could not thank you for such a detailed explanation!!! I am convinced now. My experience here at mathstack exchange has been great! Thank you again! Cheers from Brazil! $\endgroup$ – Lucas Oct 30 at 5:04
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The short answer is: It depends on the situation.

Here, you can't choose $p=1$, as it is given that $p$ is a fixed natural number.

However, when it comes to working with $\epsilon$'s, you are required to show a result to be true for all $\epsilon > 0 $. Hence, given an $\epsilon > 0$, its a convention for authors to use, "choose $\epsilon = \frac{\epsilon}{2}$ ", when they actually mean to say, they are trying to prove the result for this fixed $\epsilon$, by choosing some ${\epsilon}' = \frac{\epsilon}{2} $, and thus, they are proving the result for all $\epsilon > 0$, as this fixed $\epsilon$ was arbitrary.

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No, you cannot take $p=1$, in order to prove the statement you need to show that it holds for arbitrary $p \in \mathbb{N}$. Taking $\epsilon' = \frac{\epsilon}{2}$ is allowed because this still works for an arbitrary choice of $\epsilon$. If I take $\epsilon = 0.1$ or if take it to be $\epsilon =0.9$, the proof still works since $\epsilon'$ is defined relatively, but if you only prove something for say $p=1,2$, then you have not proven the claim generally, and your proof might break down for $p=4$ say

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  • $\begingroup$ But what does it mean to work for an arbitrary choice of $\epsilon$? It always looks to me as if we are proving the claim only for $\frac{\epsilon}{2}$ only $\endgroup$ – Lucas Oct 30 at 4:39
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    $\begingroup$ why does this make a difference though? If i said take $\epsilon=0.1$, then it would be an issue, but if take $\epsilon'=\epsilon/2$, then the proof still works for any choice of $\epsilon>0$ right? As in, for any $\epsilon>0$ you choose, the proof will always work out because you can always also find a corresponding $\epsilon'$ $\endgroup$ – dimebucker Oct 30 at 4:43
  • $\begingroup$ So can i take, for instance, $p^{\prime} = \frac{p}{2}$ and prove the statement? $\endgroup$ – Lucas Oct 30 at 4:50
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    $\begingroup$ well p/2 might no longer be an integer, so this might not be useful.. but you will see some proofs where they might break it into two cases, $p=2k$ and $p=2k+1$ for some integer $k$ (even and odd cases).. $\endgroup$ – dimebucker Oct 30 at 4:59
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Well, usually the word "let" it's use for defining the context of a variable and also for letting fixed its value. However, it's usual to redefine the value of certain variable as in your example "$\epsilon = \frac{\epsilon}{2}$" strictly speaking, it shouldn't happen, but it's a convention, in order to avoid choosing from the beginning the value $\epsilon/2$ or others, which works but which could be seen as non intuitive, and to avoid rewrite the whole proof from the start.

All you know about $p$, it's that is a natural number, could be any, but you cannot choose one at your convenience, unless you will go into all of them, for example considering $p$ as an odd number, then as an even number, and so on...

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You have no choice about $p$: you have to find an answer that works no matter what $p$ you are given. For example, $$x_n=\cos\frac{2\pi n}p$$would work for any given nonzero $p$, while $x_n=n$ would do in the trivial case $p=0$.

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