Exercise 3.23

Question

Exercise 23:
    Suppose $\{p_n\}$ and $\{q_n\}$ are Cauchy sequences in a metric space $X$. Show that the sequence $\bigl\{d(p_n,q_n)\bigr\}$ converges.

ghostofgarborg · Accepted Answer

\def\eps{\varepsilon} By applying the triangle inequality, we see that $$ d(p_n, q_n) - d(p_m,q_m) \leq d(p_n, p_m) + d(q_m, q_n) $$ By interchanging the roles of $m$ and $n$, we in fact get $$ |d(p_n, q_n) - d(p_m,q_m)| \leq d(p_n, p_m) + d(q_m, q_n) $$ Let $\epsilon > 0$. Since $p_n$ and $q_n$ are Cauchy, we can find an $N$ such that $m, n > N$ implies that $d(p_n, p_m) < \frac \epsilon 2$ and $d(q_m, q_n) < \frac \epsilon 2$. Then $$ |d(p_n, q_n) - d(p_m,q_m)| \leq \epsilon $$ This shows that $d(p_n, q_n)$ is Cauchy, hence convergent.

Exercise 3.23

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Exercise 3.23

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