Proof that Cauchy in $\mathbb{R}^2$ implies convergence.

Question

suppose $\{(x_n,y_n)\}$ is a Cauchy sequence in $(R^2,d_2)$. prove $\{(x_n,y_n)\}$ converges in $R^2$.

let $\epsilon>0$ and $S_n=\{(x_n,y_n)\}$ then there exists $N\in I$ s.t $d_2(S_n,S_m)<\sqrt\epsilon$ for all $n,m\ge N$

this implies$$(|x_n-x_m|^2+|y_n-y_m|^2)^\frac{1}{2}<\sqrt\epsilon$$ so,$$(|x_n-x_m|^2+|y_n-y_m|^2)<\epsilon$$ which implies$$|x_n-x_m|^2<\epsilon$$ and $$|y_n-y_m|^2<\epsilon$$ I am going to conclude ${x_n}$ , ${y_n}$ are Cauchy( but I can not conclude this for the last two inequalities, right? since what if the squared absolute value is between 0 and 1?) , hence convergent in $R^1$and then conclude $S_n$ converges in $R^2$ but I do not know how to do this.

Answer 1

Given $\epsilon > 0$, there exists $N$ such that $d(S_n, S_m) < \epsilon$ for $n,m \ge N$. Then $|x_n - x_m| \le d(S_n, S_m) < \epsilon$ holds for $n,m \ge N$. Thus $(x_n)$ is a Cauchy sequence.

Answer 2

I can see you're concerned about taking the square root at the end of the problem there.

You would have saved yourself that concern if, instead of starting out with that $\sqrt{\epsilon}$ inequality, you had started out with an $\epsilon$ inequality instead, namely $d_2(S_n,S_m) < \epsilon$.

That would have implied $$(|x_n-x_m|^2+|y_n-y_m|^2)<\epsilon^2 $$ which implies $$|x_n-x_m|^2<\epsilon^2 \quad\text{and}\quad|y_n-y_m|^2<\epsilon^2 $$ and now its safe to take the square root on both sides of each of these inequalities, using that the square root function is strictly monotonic.