Proving $ \lim_{n \to \infty} a_n = \lim_{n \to \infty} b_n $ Cantor's Lemma

Question

In Cantor's Lemma, we prove $ \lim_{n \to \infty} a_n = \lim_{n \to \infty} b_n $ in the following way:

Let $ \lim_{n \to \infty} \, a_n = L $

$ \lim_{n \to \infty} \, b_n = \lim_{n \to \infty} \, (b_n - a_n) + a_n = \lim_{n \to \infty} \, b_n - a_n + \lim_{n \to \infty} \, a_n = 0 + L = L$

But why can't we just calculate it like this:

$ \lim_{n \to \infty} \, b_n - a_n = 0 \Rightarrow \lim_{n \to \infty}\, b_n - \lim_{n \to \infty} \, a_n = 0 \Rightarrow \lim_{n \to \infty} \, b_n = \lim_{n \to \infty} \, a_n$

The Lemma: Let $ a_n, b_n $ be sequences such that $ \forall n, \, a_n \le a_{n + 1} \le b_{n + 1} \le b_n, \, \lim_{n \to \infty} (b_n - a_n) = 0$

$ a_n $ is bounded above by $ b_1 $ and $ \forall n, a_{n + 1} \ge a_n $. Therefore $ a_n $ is convergent.

$ b_n $ is bounded below by $ a_1 $ and $ \forall n, b_{n + 1} \le b_n $. Therefore, $ b_n $ is convergent.

Let $ \lim_{n \to \infty} a_n = L $

Therefore:

$ \lim_{n \to \infty} b_n = \lim_{n \to \infty} (b_n - a_n + a_n) = \lim_{n \to \infty} (b_n - a_n) + \lim_{n \to \infty} a_n = 0 + L = L $

Answer 1

Let $ \lim_{n \to \infty} \, a_n = L $

$ \lim_{n \to \infty} \, b_n = \lim_{n \to \infty} \, (b_n - a_n) + a_n = \lim_{n \to \infty} \, b_n - a_n + \lim_{n \to \infty} \, a_n = 0 + L = L$

But why can't we just calculate it like this:

$ \lim_{n \to \infty} \, b_n - a_n = 0 \Rightarrow \lim_{n \to > \infty}\, b_n - \lim_{n \to \infty} \, a_n = 0 \Rightarrow \lim_{n \to > \infty} \, b_n = \lim_{n \to \infty} \, a_n$

You certainly can do it either way. But since both ways are equally long, I don't see why the second would be inherenly better than the first.

In fact, I think the first way is better for at least two reasons:

(1) Elegance. The first way is a proof that consist entirely of equalities, that is it is of the form $\lim_{n\to\infty} b_n = ... = ... = ... = \lim_{n\to\infty}a_n$, which is a very elegant form of a proof.

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(2) Lack of hidden assumptions. Every equality in the first proof is completely justified by all the facts that are known beforehand:

1. $\lim_{n \to \infty} \, b_n = \lim_{n \to \infty} \, (b_n - a_n) + a_n$ is true because $b_n = (b_n - a_n) + a_n$ is true.
2. $\lim_{n \to \infty} \, (b_n - a_n) + a_n = \lim_{n \to \infty} \, b_n - a_n + \lim_{n \to \infty} \, a_n$ is true because the limit of a sum is the sum of limits, if the individual limits exist, and we already know the two sublimits exist.

In comparison, in your version of the proof, you assume that $\lim_{n \to \infty} \, b_n - a_n = \lim_{n\to\infty}\, b_n - \lim_{n \to \infty}$, but this is only true if the limit of $b_n$ exists, and this (while being true) is not explicitly stated. That makes the second proof less rigorous.