Difficulty understanding manipulation of $\epsilon$ and $N$ in the definition of a limit in proofs

Question

The definition I am working with is from Foundations of Mathematical Analysis by Johnsonbaugh and Pfaffenberger.

Def: Let $\{ a_n \}_{n=1}^\infty$ be a sequence of real numbers. We say that $\{ a_n \}_{n=1}^\infty$ has limit $L\in \Bbb R$ if for every $\epsilon > 0$, there exists a positive integer $N$, such that if $n \geq N$, then

$$|a_n - L|<\epsilon$$

In the proof of the theorem stating that the sum of the the two convergent sequences is convergent to the sum of their limits a simple strategy is to take the epsilon for each sequence individually and use the definition and divide the epsilon by two so that using triangle inequality you can say:

$$|(a_n + b_n)-(L+M)| \leq |a_n - L| + |b_n - M| < \epsilon/2 + \epsilon/2 = \epsilon$$

I don't understand why this is okay to do, I kind of understand the manipulation of $N$ so that $n$ holds the same inequality of the definition but something seems logically flawed in the assertion above and it is used often for a variety of limit proofs, particularly in proving basic algebra of limit proofs.

Answer 1

By definition, if you give me a convergent sequence $(a_n)$ and any positive real number $\epsilon$, I can give you a threshold $N$ such that $a_n$ is $\epsilon$-close to its limit whenever $n \geq N$.

Therefore, if we want to prove that $(a_n) + (b_n)$ converges to the sum of the respective limits of $(a_n)$ and $(b_n)$, we must fix an arbitrary positive $\epsilon$ and find a threshold $N$ such that $a_n + b_n$ is $\epsilon$-close to that sum whenever $n \geq N$.

Now, $\epsilon$ is a fixed number. Therefore $\epsilon/2$ is another fixed positive real number. By the definition of the convergence of $(a_n)$, applied to the number $\epsilon/2$, we can find a threshold $N_1$ such that $a_n$ is $\epsilon/2$-close to its limit whenever $n \geq N_1$. Similarly, we can find a threshold $N_2$ such that $b_n$ is $\epsilon/2$-close to its limit whenever $n \geq N_2$. Taking $N = \max(N_1,N_2)$ and invoking your estimate concludes the proof.

Answer 2

Assume some sequence $\{ a_n \}_{n=1}^\infty$ has a limit $L\in \Bbb R$. Then we can pick some fixed real number greater than zero which we call $\epsilon$, and assert there exists a positive integer $N$ dependent on $\epsilon$, so we could write this $N$ in functional notation as $N(\epsilon)$, such that if $n \geq N(\epsilon)$, then

$$|a_n - L|<\epsilon\tag{1}$$

In these proofs $\epsilon$ is a dummy variable that is some fixed real number greater than zero, that we don't in general need to know explicitly numerically, since we then extend the concept to all $\epsilon>0$ in the limiting process of convergence which follows since our $\epsilon$ was arbitrary, and so can be taken as small as we like. We could equivalently state the above limiting process in terms of some other number $\xi=\frac{\epsilon}{2}$. What we then have to do is play the $\epsilon$, $N(\epsilon)$ game again with $\xi$, $N(\xi)$, and find an$N(\xi)$, such that if $n \geq N(\xi)$, then

$$|a_n - L|<\xi$$

Here nothing is broken in the definition of convergence - you still have a positive real number $\xi$ that bounds the difference $|a_n - L|$ for all $n\ge N(\xi)$.

For two convergent sequences you want the analogous result to (1): $$|(a_n + b_n)-(L+M)| <\epsilon$$ so you can say $\{a_n +b_n\}_{n=1}^\infty$ has a limit $L+M\in \Bbb R$. To do this you just do the same thing that you did with one convergent sequence to the two sequences individually after splitting them up with the triangle inequality; that is given an $\epsilon_1>0$ such that for all $n\ge N(\epsilon_1)$ we have $|a_n - L|<\epsilon_1$, and given an $\epsilon_2>0$ such that for all $n\ge N(\epsilon_2)$ we have $|b_n - M|<\epsilon_2$, and we do this in such a manner that $\epsilon_1+\epsilon_2=\epsilon$. Now pick $N=\max(N(\epsilon_1),\,N(\epsilon_2))$; we do this since if $N(\epsilon_1)<N(\epsilon_2)$, then, for all $n\ge N(\epsilon_1)$ we have $|a_{n} - L|<\epsilon_1$, but if $n<N(\epsilon_2)$ we have $|b_{n} - M|>\epsilon_2$, picking $N$ solves this. So for all $n\geq N$ we have

$$|(a_n + b_n)-(L+M)| \leq |a_n - L| + |b_n - M| < \epsilon_1 + \epsilon_2 = \epsilon$$

Note there is nothing special about $\epsilon/2$ proofs, we usually pick these as it makes it easier when dealing with two convergent sequences to just half the $\epsilon$, then you have for $n\geq N(\frac{\epsilon}{2})$

$$|(a_n + b_n)-(L+M)| \leq |a_n - L| + |b_n - M| < \epsilon/2 + \epsilon/2 = \epsilon$$ which proves convergence since $\epsilon$ was arbitrary. Note how you fix your $\epsilon$ and then work around it i.e., one minute you look at $|a_n - L|<\epsilon$ for $n\ge N(\epsilon)$, then you may need to switch to $|a_n - L|<\frac{\epsilon}{2}$ for $n\ge N(\frac{\epsilon}{2})$, but the $\epsilon$ is the same in each case.

Answer 3

I think the slight abuses of notation are what's tripping you up.

What they do here is a couple of things: First, they let $\epsilon>0$ as usual, but then they define what I will denote as $\epsilon_0$ to be half of $\epsilon$. Surely, $\epsilon_0$ will still be greater than $0$, hence it is a perfectly valid choice of an 'epsilon' for each of the two sequences individually. Since those sequences are each convergent, then they have $N_a$ and $N_b$ respectively, such that $|a_n-L|<\epsilon_0$ and $|b_n-M|<\epsilon_o$ respectively. Moreover, from the definition of limit, we note that those inequalities hold for all $n>N_a$ for $a_n$ and all $n>N_b$ for $b_n$, hence those inequalities hold for both of them for all $n>N=\operatorname{max}\{N_a,N_b\}$.

Therefore, we have that $|a_n-L|+|b_n-M|<2\epsilon_0=\epsilon$ holding for all $n>N$ as required, and the first inequality, $|(a_n+b_n)-(L+M)|\leq|a_n-L|+|b_n-M|$ comes from the triangle inequality.