Why manipulation of $\epsilon$ is allowed in theorems related to limits in real analysis?

Question

I am further explaining my question, and for that, I shall use over-simplified and imaginary scenarios for the sake of the discussion.

If $a<c$ and $b<c$, then we have $a+b<2c$. If $c>0$, then $c/2>0$. But from the first two conditions, we do not know for sure if $a<\frac{c}2$, but for sure $a<2c$.

Similarly in the definition of convergence, we have $|x_n-x|<\epsilon$, the terms are according to the definition. But we do not know for sure $|x_n-x|<\frac{\epsilon}2$ is true or not. Just like we are not sure that, for $n<K(\epsilon)$, whether the quantity $|x_n-x|$ is always less than epsilon or not.

Then in some theorems, when it is applicable, we see that sometimes we write it as granted or given that $|x_n-x|<\frac{\epsilon}2$, but no such condition is stated in the theorem or the claim, and from this stage, we reach to our final, required conclusion, where we have $\epsilon=\frac{\epsilon}2+\frac{\epsilon}2$, or $\epsilon=M\left(\frac{\epsilon}{2M}+\frac{\epsilon}{2M}\right)$ which matches with our definition.

My questions are,

1. Why, by which mathematical rule, can we manipulate $\epsilon$ in such a way?
2. Some people write $\epsilon=\frac{{\epsilon}'}2$. If $\epsilon$ here is just a synonym for 'a positive number', preferably small, then what is wrong with getting $2\epsilon$?
3. If it is absolutely necessary that the final result has 'a free epsilon' to prove our required convergence, when we state the given convergences, we are not writing those with 'free' $\epsilon$'s, so it's a paradox from the logic 'we can not prove convergence without a free epsilon'.

On a side note, in $K(\epsilon)$ game, player A claims a quantity to be the limit for a certain sequence, and player B challenges A by giving him a certain $\epsilon$. Then player A has to prove convergence with that particular $\epsilon$ to win, id est, to prove that he has the right quantity for limit. But as far as I understand, there can be infinite number of such epsilons which player B can provide, and thus, the game continues for eternity, there is no way to tell that the proposed limit is the right limit according to the definition in this way. So how does this process help with finding the limit of a convergent sequence using the definition? What am I missing?

Answer 1

Expanding on my comment, I just want to mention that your points 2,3 are correct. If $\epsilon$ represents all positive numbers then $2\epsilon $ also represents all positive numbers and it is not necessary to have just $\epsilon$ in the final part of a proof. That many books and instructors (and even myself) do it is more due to convention and not because of some mathematical necessity.

Also your last paragraph about the game being played between $\epsilon$ and $\delta$ is very true. But remember that it is a hypothetical game and one has to deal with all positive $\epsilon$ and not just a finite number of chosen $\epsilon$. So the game analogy is only meant to help you in understanding the concept. In your proofs you should start with the symbol $\epsilon$ and make no assumptions about it except $\epsilon >0$ and proceed accordingly. That takes care of all $\epsilon>0$. Never try to substitute actual numbers like $\epsilon=0.0001$ in your argument.

Also understand that the processes of analysis / calculus mostly deal with infinite/infinity and you need to get accustomed to it. The occurrence of infinite is common in other branches of mathematics also but it is not stressed out so directly like in analysis. For example when we state that $a+b=b+a$ ie addition is commutative, we are actually saying that the identity holds for all real $a, b$ and the situation is similar to "all $\epsilon>0$". Contrary to what many believe the real problem in understanding these definitions of analysis is not the use of logical quantifiers like $\forall, \exists$ but rather the statements involving inequalities. No one ever got disturbed by the use of "for all" in the statement $$ a+b=b+a\text{ for all real numbers }a, b$$ from algebra.

Also note that the definition of limit is not an effective / practical tool to find /evaluate a limit. Rather the definition is used to prove simple yet powerful limit theorems which are then used to evaluate limits. The point of typical $\epsilon, \delta $ exercises is not to teach you yet another technique of finding limits, but they are designed to help you understand the definition of limit (however I doubt if these exercises really help in that direction because the focus shifts from the definition to the algebraic problem of finding $\delta, K(\epsilon) $ explicitly in terms of $\epsilon$).

Answer 2

Let's say, for sake of example, that we are going to prove that $\lim_{n\to\infty}(x_n+y_n)=\lim_{n\to\infty}x_n+\lim_{n\to\infty}y_n$, assuming both the limits on the right exist. Set them equal to $x$ and $y$, respectively.

To prove this, we take an arbitrary $\epsilon>0$, and we show that there is a natural number $K(\epsilon)>0$ such that for all $n>K(\epsilon)$, we have $$|x_n+y_n-(x+y)|<\epsilon$$The only thing we have to work with is that $x_n$ and $y_n$ converge to their respective limits, and that $$ |x_n+y_n-(x+y)|\leq|x_n-x|+|y_n-y| $$ If we can force each of the right-hand terms to be less than $\epsilon/2$, then we're good to go. And we can! Because they converge, we can force them to be smaller than any positive number. When they are the main limits we care about, we usually call that number $\epsilon$, but this time they aren't the main limit, we already have an $\epsilon$, and the main thing we care about is that these terms are small enough compared to $\epsilon$. Therefore we choose to make them smaller than $\epsilon/2$.

There are many different ways to phrase this, but the general idea behind it is always the same.