1. ## Check my proof. Please :)

So I'm studying analysis on my own which has the disadvantage that there aren't many people capable of checking proofs in my neighborhood. So if you would be so kind, I would truly appreciate it!

Problem:
Suppose that $\displaystyle \{a_n\}$ is a Cauchy sequence. Prove that $\displaystyle \{a_{n}^{2}\}$ is a Cauchy sequence.

I came up with two proofs.

Proof 1:
Since $\displaystyle \{a_n\}$ is a Cauchy sequence, and the real numbers are complete, $\displaystyle \lim_{n\rightarrow \infty}a_n=a$ for some $\displaystyle a\in\Re$. Since the limit of a product of convergent sequences is the product of the limits of the convergent sequences, we have

$\displaystyle \lim_{n\rightarrow \infty}$$a_{n}^{2}$$=$${\lim_{n\rightarrow \infty}a_{n}}$$^2=a^2$.

Since $\displaystyle \{a_{n}^{2}\}$ converges to a finite limit, it is a Cauchy sequence. QED.

Proof 2:
Since $\displaystyle \{a_n\}$ is a Cauchy sequence, it converges to a finite limit. Since it converges to a finite limit, it is bounded. That is, there exists a number $\displaystyle M$ so that $\displaystyle \mid a_n\mid \leq M$ for all $\displaystyle n$. Using the triangle inequality, we see that

$\displaystyle \mid a_n+a_m \mid \quad \leq \quad \mid a_n\mid+\mid a_m \mid \quad \leq \quad 2M$ for all $\displaystyle m$ and $\displaystyle n$.

Let $\displaystyle \varepsilon > 0$ be given. Since $\displaystyle \{a_n\}$ is a Cauchy sequence, there exists an $\displaystyle N$ so that $\displaystyle n\geq N$ and $\displaystyle m\geq N$ implies $\displaystyle \mid a_n-a_m \mid \leq \frac{\varepsilon}{2M}$. But this means that
$\displaystyle \mid a_n^2-a_m^2\mid \quad = \quad \mid a_n+a_m \mid \mid a_n-a_m\mid \quad \leq \quad 2M\mid a_n-a_m \mid \quad \leq \quad \varepsilon$.

Therefore, $\displaystyle \{a_n^2\}$ is a Cauchy sequence. QED.

2. ## Re: Check my proof. Please :)

Both of your proofs assume that you have a sequence of real numbers, but that isn't specified in the problem.

In proof 2, you only use the completeness of the real numbers to show that the sequence is bounded. I think you can show that without assuming completeness.

- Hollywood

3. ## Re: Check my proof. Please :)

Both of your proofs assume that you have a sequence of real numbers, but that isn't specified in the problem.

In proof 2, you only use the completeness of the real numbers to show that the sequence is bounded. I think you can show that without assuming completeness.

- Hollywood
Hi Hollywood,

Thank you for taking time to look at my proofs. I really appreciate it.

I guess I should have prefaced the post with the fact that I'm studying real analysis, so it is implied that I'm dealing with a sequence of real numbers.

In proof 2, I used completeness to show the sequence converges. Which in turn I used to show the sequence is bounded.