Using Cantor's lemma in a proof of convergence

Question

I was given the following question: Let $q$ be a real number such that $0 < q < 1$, and let ${a_n}$be a sequence that follows the rule : $|a_{n+1}-a_n| < q|a_n-a_{n-1}|$ for any $n>1$. Prove that ${a_n}$ converges.

I already know this can be proven using Cauchy's criterion, but I had a different proof in mind (using Cantor's lemma), and wanted to know if it's valid, or if I might be mistaken. So here it is:

First - I prove that the sequence is bounded. Since the sequence is bounded, the set of its members (and the set of any sub sequence of ${a_n}$) has a Supremum and an Infimum. Then, for any natural number $k$, I define the closed interval $A_{k}$, such that $A_{n}=[\inf(a_k), \sup(a_k)]$, where inf and sup of $k$ are the Infimum and Supremum of the set of the members of $a_n$, starting from $a_k$ (meaning, any member with an index greater than or equal to $k$). Naturally, $A_1$ is the "biggest" interval, and because of how the sequence is defined, $A_{n+1}$ is always a proper subset of $A_n$ for any $n$. Now I use Cantor's lemma - because the interval gets infinitely smaller and smaller (after n iterations, it gets at least $q^n$ times smaller), the intersection of all intervals $A$ (meaning, $A$ of $n=\infty$) contains one single point, let's call it $L$. Now, I state that $L$ must be the limit of $a_n$ - for any epsilon, there exists a natural number $k$, such that for any $n>k$, $A_k$ is contained in the epsilon neighborhood of$ L$, and therefore all members of $a_n$ starting from $a_k$ are in the epsilon neighborhood of $L$, which means $L$ must be the limit.

So do you think this is a valid proof? Or am I making some false assumptions? I know this proof is a bit longer than using Cauchy's criterion, but I like it better because I feel like it gives a better explanation as to 'why' the series converges. Thanks in advance!

Answer 1

It is a valid proof - at least with the addendum of the argument JonathanZ has pointed out is missing: that the interval sizes go to $0$.

Your notation leaves something to be desired. Sets have infimums and supremums, not individual elements. You could have defined $A_k = \{a_i \mid i \ge k\}$ and then called them $\inf A_k$ and $\sup A_k$, or just called them $\inf \{a_i \mid i \ge k\}$ or even $\inf_{i \ge k} a_i$. Either one would be acceptable.

You also don't need Cantor's lemma - you've already made use of the completeness of $\Bbb R$ in claiming the existance of $m_k = \inf \{a_i \mid i \ge k\}$ and $M_k = \sup \{a_i \mid i \ge k\}$.

Since for all $i \ge k, m_k \le m_i \le a_i \le M_i \le M_k$, the sequence $\{m_i\}_{i=1}^\infty$ is increasing and bounded above by all $M_k$, while $\{M_i\}_{i=1}^\infty$ is decreasing an are bounded below by all $m_k$. Therefore $$\lim_{k \to \infty} m_k = \sup \{m_i\}_{i=1}^\infty = \liminf_k a_k \le \limsup_k a_k = \inf \{M_i\}_{i=1}^\infty = \lim_{k\to\infty} M_k$$

Therefore, your argument can be simplified to:

Since $\{a_k\}_{k=1}^\infty$ is bounded, $\liminf_k a_k$ and $\limsup_k a_k$ exist and since $\sup_{i \ge k} a_i - \inf_{i \ge k} a_i\to 0$ as $k \to \infty$, we have $\liminf_k a_k = \limsup_k a_k$. Therefore $\lim_k a_k$ converges to the common value.

Add actual arguments showing that $a_k$ is bounded and that $\sup_{i \ge k} a_i - \inf_{i \ge k} a_i\to 0$, and you are done.

Answer 2

I'm afraid there's a fatal flaw in your proof, and it lies in your claim that $$\text{after } n \text{ iterations, it gets at least } q^n \text{times smaller} \tag{1}.$$ It turns out that we can't deduce this from $\lvert{a_{n+1}} - a_{n}\rvert < q*\lvert{a_{n}} - a_{n-1}\rvert$, (which I'll call '$q$-contracting'). Here's a very basic counter-example:

$$(0, 1, 1/2, 1/4, 1/8, ....).$$

It's $q$-contracting for any $q \gt 1/2$, but your interval $A_n$ stays the same size on the first iteration. And while I haven't does the work, it seems like it would be possible to put scaled-down versions of this all the way out to infinity, so we wouldn't even be able to prove that $(1)$ holds "eventually".

So you'd have to find another way to prove that the length of the intervals goes to $0$. But note the following: because we know that $\{a_i\}$ is bounded above and below it's not too hard to show that $$ \lvert A_n\rvert = \sup_{i>n}\{a_i\} - \inf_{i>n}\{a_i\} = \sup_{i,j > n}\{ a_i - a_j\} = \sup_{i,j > n}\{\lvert a_i - a_j\rvert\},$$

so proving that $\lim_{n \rightarrow \infty} \lvert A_n\rvert = 0$ means proving that that second supremum goes to $0$ as $n \rightarrow \infty$, i.e. $\forall \epsilon > 0\ \exists N$ such that $\lvert a_i - a_j\rvert \lt \epsilon$ for all $i,j \gt N$. In other words, proving that the length of your intervals goes to zero is basically the same as establishing that the Cauchy condition holds for your sequence.

Since it feels to me like you may be interested in a more general critique, I'll add that I think what happened is you recast the problem in the context of those decreasing intervals, and in doing so the ugly details got hidden under your assertion $(1)$. It's great if your intervals version gives you more insight into what's going on, but it looks like you'll still have to do the same grunge work as the Cauchy criterion version.