Analysis

**Mathstud28** · November 29th 2008, 02:53 PM

Hello everyone. I am posting this not because the normal proof is difficult, but because I think I have come up with an alternative way of doing it and am wondering if I made any fatal errors. Also, half the reason I am putting this up here is to improve my often remarked upon lack of coherency. So any constructive criticism in either category is much appreciated.

Definitions: If $\chi$ is a metric space and $\left\{p_n\right\}$ is a sequence under $\chi$ , then we define an infinite subsequence of $\left\{p_n\right\}$ as follows: If we have a infinite sequence of natural numbers $n_1,n_2,\cdots,n_k$ where $n_1<n_2<\cdots<n_k$ then the sequence $\left\{p_{n_k}\right\}$ is a subsequence of $\left\{p_n\right\}$ .

So now that we have defined an infinite subsequence, let's get to the question.

Question: If $\left\{p_n\right\}$ is a convergent sequence under $\chi$ which converges to $p$ then prove that every infinite subsequence $\left\{p_{n_k}\right\}$ of $\left\{p_n\right\}$ converges to $p$ .

Answer: I think I know how they intended it to be done, but I came up with a different method and would appreciate any crticism.

1. In the definition of the subsequence we defined the $n_k$ in $\left\{p_{n_k}\right\}$ to be sequence of natural numbers such that $n_1<n_2<\cdots<n_k$ . So instead of $n_k$ let us define it as follows: let $n_k$ be represented by $\varphi: \mathbb{N}\mapsto\mathbb{N}$ which is a function of $n$ and has the quality that it is monotonically increasing, so then we may represent a subsequence of $\left\{p_n\right\}$ as $p_{\varphi}$ .

2. We know because $\left\{p_n\right\}$ converges that for every $\varepsilon>0$ there exists and integer $N$ such that $\forall{N\leqslant{n}}~d_{\chi}\left(p,p_n\right)< \varepsilon$ .

3. So from this we can make the observation that $d_{\chi}\left(p,p_{f}\right)<\varepsilon$ whenever $N\leqslant{f^{}\left(n\right)}\implies{\lceil{\til de{\varphi}\left(N\right)\rceil}}=N_0\leqslant{n}$ . Where $\tilde{\varphi}$ is the generalized inverse function.

4. Stating the end part of 4. more explicitly: for every $\varepsilon>0$ there exists an integer $N_0$ such that $d_{\chi}\left(p,p_f\right)<\varepsilon$ whenever $N_0\leqslant{n}$ . This the definition of a sequence which converges to the value $p$ which is what was asked to be proven.

Any comments greatly appreciated,
Mathstud.

**Laurent** · November 29th 2008, 03:12 PM

Originally Posted by Mathstud28

2. So now here is where I might have a problem if there is one.

Based on how we defined $f$ above it is clear that $f^{-1}$ exists.

You're right, that's where you have a problem. $f$ is increasing, hence $f$ is one-to-one, but it not onto, so you can't define $f^{-1}(N)$ for any $N$ .

You can patch this by saying: let us define the "generalized inverse function" $\widetilde{f}$ by $\widetilde{f}(N)=\min\{n\geq 0\,|\,f(n)\geq N\}$ . By doing so, you still can say:

3. We know because $\left\{p_n\right\}$ converges that for every $\varepsilon>0$ there exists and integer $N$ such that $\forall{N\leqslant{n}}~d_{\chi}\left(p,p_n\right)< \varepsilon$ .

4. So from this we can make the observation that $d_{\chi}\left(p,p_{f}\right)<\varepsilon$ whenever $N\leqslant{f^{}\left(n\right)}\implies\widetilde{f }(N)=N_0\leqslant{n}$ .

and it works.

But honestly, it seems to me that this is exactly the usual proof, written a bit differently... (with an explicit $N_0$ )

**Mathstud28** · November 29th 2008, 03:13 PM

Originally Posted by Laurent

You're right, that's where you have a problem. $f$ is increasing, hence $f$ is one-to-one, but it not onto, so you can't define $f^{-1}(N)$ for any $N$ .

You can patch this by saying: let us define the "generalized inverse function" $\widetilde{f}$ by $\widetilde{f}(N)=\min\{n\geq 0\,|\,f(n)\geq N\}$ . By doing so, you still can say:

and it works.

But honestly, it seems to me that this is exactly the usual proof, written a bit differently...

Hmm, maybe it is the standard proof, but not the one my book supplies. Thank you for your time Laurent!

EDIT: After you said something I looked back and realized I was thinking of the wrong proof, my book "Principles of Mathematical Analysis" omits this proof. So I'm kind of happy that it is the common proof...all be it with the wrong notation

**Opalg** · November 30th 2008, 12:11 AM

Originally Posted by Laurent

You're right, that's where you have a problem. $f$ is increasing, hence $f$ is one-to-one, but it not onto, so you can't define $f^{-1}(N)$ for any $N$ .

It's true that f does not have an inverse, but since it is injective it does have a left inverse, which is all that is needed for this argument. Personally, I found Mathstud28's suggestion to be a succinct version of a proof that is sometimes presented in a much less clear way.

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