Cauchy sequence real analysis

Question

Prove that if a sequence $X_n$ is Cauchy, then the sequence $X^2_n$ is Cauchy.

Answer 1

Lemma: A Cauchy sequence $a_1,a_2,a_3,...$is bounded. Proof: Let $\nu$ be such that $|a_s-a_t|<1$ whenever $s,t \ge \nu.$ Thus, for $s \ge \nu,$ $$|a_s-a_{\nu}|<1$$ $$a_{\nu}-1<a_s< a_{\nu}+1$$ $$|a_s|<\max(|a_{\nu}-1|,| a_{\nu}+1|)$$ $$\text {Let }K=\max(|a_1|,…,|a_{\nu -1}|,|a_{\nu}-1|,| a_{\nu}+1|)$$ Thus $|a_s| \le K$ for all $s$, proving the lemma.Now we can prove that the sequence of squares is also Cauchy. Let $\epsilon>0.$ Let $p$ be such that $|a_m-a_n|<\frac{\epsilon}{2K}$wnever $m,n \ge p.$. Then, for $m,n \ge p,$ $$|a_m^2-a_n^2|=|a_m-a_n||a_m+a_n|$$ $$\le |a_m-a_n|(|a_m|+|a_n|)$$ $$<\frac{\epsilon}{2K}(K+K)=\epsilon,Q.E.D.$$