# “Let” in math texts

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.

• 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 $p$. – littleO Oct 30 at 4:54
• Does Bartle actually say "Let $\epsilon=\epsilon /2$" after stating that $\epsilon>0$? – DanielWainfleet Oct 30 at 5:04
• @DanielWainfleet: If Bartle does that, then Bartle is wrong. That's just nonsense. Please, Lucas, clarify whether Bartle really does that. – user21820 Oct 30 at 16:13
• And I strongly suggest you read carefully through this post about "let". – user21820 Oct 30 at 16:14
• 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 – Lucas Oct 30 at 16:44

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 $$C = 1 + \epsilon^2$$.

• 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 $\epsilon$ – Lucas Oct 30 at 4:41
• @Lucas: I updated my answer a little bit to furnish more examples and also address "the epsilon issue." Hopefully it helps! – Alex Ortiz Oct 30 at 4:54
• 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! – Lucas Oct 30 at 5:04

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.

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

• 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 – Lucas Oct 30 at 4:39
• 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 $\epsilon'$ – dimebucker Oct 30 at 4:43
• So can i take, for instance, $p^{\prime} = \frac{p}{2}$ and prove the statement? – Lucas Oct 30 at 4:50
• 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).. – dimebucker Oct 30 at 4:59

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...

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 $$p=0$$.