Talk:Sequence of Powers of Number less than One/Necessary Condition/Proof 4

Concerning Proof 4, I want to raise the issue that $\ds \lim_{n\to\infty} x_n = a$ implies $\ds \lim_{n\to\infty} f(x_n) = f(a)$ for continuous functions. However, the converse does not generally hold, for example when $f$ is identically zero everywhere. It appears to me as Proof 4 relied on the converse. Comments? --Lord_Farin 12:32, 9 March 2012 (EST)


 * Proof 4 does not rely on the converse. The proof shows that $0 = \ds \lim_{n\to\infty} y_n = \lim_{n\to\infty} \left\vert x^n \right\vert = \left\vert \lim_{n\to\infty} x^n \right\vert$, hence $\ds \lim_{n\to\infty} x^n = 0$. Abcxyz 13:54, 9 March 2012 (EST)


 * Actually, it would, weren't it the case that the modulus is injective to $0$, i.e. $\left\vert{x}\right\vert = 0\implies x =0$. But that's an axiom for the modulus function and is different from the continuity assertion (compare indeed when $f$ is the zero function). You may want to think a bit on this. --Lord_Farin 19:11, 9 March 2012 (EST)


 * The continuity of the complex modulus function and the Composite of Continuous Mappings is Continuous are used in the equality $\ds \lim_{n\to\infty} \left\vert x^n \right\vert = \left\vert \lim_{n\to\infty} x^n \right\vert$. Abcxyz 19:32, 9 March 2012 (EST)


 * Again, the point is that you can't conclude that $\ds \lim_{n\to\infty} x_n$ exists from this equation. Continuity merely means $x_n\to a \implies f(x_n)\to f(a)$. You are using $f(x_n)\to f(a) \implies x_n\to a$, which is tailored to the special case that $f(a) = 0$. Compare $x_n = (-1)^n$. Then $|x_n|\to|1|$, but not $x_n \to a$... So a little more justification is required. --Lord_Farin 19:55, 9 March 2012 (EST)


 * I used the monotone convergence theorem to prove that $\ds \lim_{n\to\infty} y_n$ exists. Then I proved that the limit is equal to 0. Since $y_n = \left\vert x^n \right\vert$, it follows that $\ds \lim_{n\to\infty} \left\vert x^n \right\vert = 0$. After that, I use the continuity of the complex modulus function and the Composite of Continuous Mappings is Continuous to conclude that $\ds \left\vert \lim_{n\to\infty} x^n \right\vert = 0$.
 * If I correctly interpreted what you just posted, that step is the problem you have. I assumed the existence of the limit $\ds \lim_{n\to\infty} x^n$. I proved that if the limit exists, then it is equal to 0.
 * However, if the limit does not exist, then (using the $\epsilon$-$\delta$ definition of a limit) there exists a $\epsilon > 0$ such that there exist arbitrarily large $n$ such that $\left\vert x^n \right\vert \ge \epsilon$. But this is impossible since $\ds \lim_{n\to\infty} \left\vert x^n \right\vert = 0$. Is that a clearer way of presenting the argument? Or is it still flawed? Abcxyz 20:12, 9 March 2012 (EST)

It appears that the argument is bound to keep giving rise to confusion. I will post a rudimentary version of a sometimes useful result (which can almost be called an observation, but nevertheless) on my sandbox page (User:Lord_Farin/Sandbox). Hopefully, this will make clear the point I'm trying to make all along. --Lord_Farin 09:33, 10 March 2012 (EST)


 * @Lord_Farin: If $f^{-1}\left({y}\right)$ contains exactly one element, then the statement posted on your sandbox does not, in general, hold. For example, let $f : \R_{\ge 0} \to \R_{\ge 0}$ be the function defined by $f\left({x}\right) = x e^{-x}$. Then $f^{-1}\left({0}\right) = \left\{ {0} \right\}$. The sequence $f\left({1}\right),f\left({2}\right),f\left({3}\right),\ldots$ converges to 0, but the sequence $1,2,3,\ldots$ does not converge to 0.


 * I did not use the statement posted on your sandbox to conclude that $\ds \lim_{n\to\infty}x^n = 0$.


 * I admit that the reasoning is not exactly clear yet. I'll have to try rewriting the proof to make it clearer. Abcxyz 10:30, 10 March 2012 (EST)

Good point you made there. It is necessary to assume that the sequence is bounded. I agree, furthermore, that it should be stipulated that the preimage is nonempty (for the preimage of $x \mapsto \exp(-x)$ of $0$ is empty; it would be $+\infty\in\overline\R$, the extended real line (to which the result, to be sure, does not apply as it can't be appropriately endowed with a metric)). I think I should retract my statement for metric spaces, it is too general. It is genuinely required to be working on a normed vector space; sorry for that (I was brought to this insight by considering trivial metrics which assume discrete values).

I still think that the phrase 'By the continuity...' tacitly uses some kind of result like this (or you reason along the lines of its proof). I will write the proof I have in mind on my sandbox page, and if you would put the exact proof of said line here or somewhere else I notice it, that'd be great. Then we can compare and close this case. Sorry to be such a pain, but I strive for clarity and rigour. --Lord_Farin 10:38, 10 March 2012 (EST)

The page Sequence of Powers of Number less than One/Proof 4 finally illuminates what you have been trying to convey all along. You reasoned along the lines of $y_n = |x^n| = |x^n-0|$ to conclude that $x^n \to 0$ as $n\to\infty$. I'm sorry for being so oblivious and hard-headed. Referenced page made the point explicit, after which it finally dawned in my mind. Nevertheless, what I put up on my sandbox may be considered the fruit of this lengthy debate. I hope it's of some use. --Lord_Farin 10:46, 10 March 2012 (EST)