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I had some issues proving this, so I relaxed the conditions to prove the statement in $\mathbb{R}$ first. Here is my attempt:

Note: I used the following theorem:

Theorem. Every Cauchy sequence of real numbers converges.

Proof. By the theorem, we know $\{p_n\}$ and $\{q_n\}$ converges since they are Cauchy. So by definition,

$(1)$ $p_n \rightarrow p$ for given $\epsilon > 0,\ \exists N_1>0$ such that $d(p_n,p)< \epsilon /2$.

$(2)$ $q_n \rightarrow q$ for given $\epsilon > 0,\ \exists N_2>0$ such that $d(q_n,q)< \epsilon /2$.

Let $N> max(N_1,N_2)$. So $\forall n \geq N$, $$d(p_n,q_n) \leq d(p_n,p) + d(p,q) + d(q_n,q)\ (*)$$ $$< \epsilon /2 + d(p,q) + \epsilon /2$$ $$\Rightarrow d(p_n,q_n) - d(p,q) < \epsilon\ (1)$$ Similarly, mult. $(*)$ by $-1$ and rearrange to get $$d(p,q) - d(p_n,q_n) > \epsilon\ (2)$$ Thus, by $(1)$ and $(2)$ we get $$|d(p_n,q_n) - d(p,q)| < \epsilon$$ Hence, $d(p_n,q_n) \rightarrow d(p,q)$. $□$

Now, how can I generalize it to any metric space $X$? For some reason, I think it cannot be done. Thank you!

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    $\begingroup$ Show that $d(p_n,q_n)$ is a Cauchy sequence of real numbers. $\endgroup$ – uniquesolution Mar 15 '17 at 16:10
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You don't want to use the fact that $p_n,q_n$ converges in your proof because it doesn't hold unless $X$ is a complete metric space. So instead, try to show that if $p_n,q_n$ are Cauchy in $X$ then $d(p_n,q_n)$ is Cauchy in $\mathbb{R}$ and then use the fact that $\mathbb{R}$ is complete so every Cauchy sequence in $\mathbb{R}$ converges. To do that, try to estimate

$$ \left| d(p_n,q_n) - d(p_m,q_m) \right| $$

in terms of $d(p_n,p_m), d(q_n,q_m)$ using the triangle inequality and then show that this is Cauchy because $d(p_n,p_m), d(q_n,q_m)$ are Cauchy.

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  • $\begingroup$ Thank you for the tip! Here is my attempt (again!): Proof. Let $\{p_n\}$ and $\{q_n\}$ be Cauchy in $X$. Then, $(1)$ $p_n \rightarrow p$ for given $\epsilon > 0,\ \exists N_1>0$ such that $d(p_n,p_m)< \epsilon /2,\ \forall n, m> N_1$. $(2)$ $q_n \rightarrow q$ for given $\epsilon > 0,\ \exists N_2>0$ such that $d(q_n,q_m)< \epsilon /2,\ \forall n, m>N_2$. Let $N> max(N_1,N_2)$. So $\forall n \geq N$, $$d(p_n,q_n) \leq d(p_n,p_m) + d(p_m,q_m) + d(q_m,q_n)\ (*)$$ $$< \epsilon /2 + d(p_m,q_m) + \epsilon /2$$ $$\Rightarrow \boxed{d(p_n,q_n) - d(p_m,q_m) < \epsilon}\ (*)_1$$ (cont.) $\endgroup$ – Curious Math Student Mar 15 '17 at 17:16
  • $\begingroup$ (cont.) Similarly, multiply $(*)$ by $-1$ and get $$-d(p_n,q_n) \geq -d(p_n,p_m) - d(p_m,q_m) - d(q_m,q_n)$$ $$ d(p_m,q_m) - d(p_n,q_n) > -\epsilon /2 - \epsilon /2$$ $$\Rightarrow \boxed{d(p_m,q_m) - d(p_n,q_n) > \epsilon}\ (*)_2$$ By $(*)_1$ and $(*)_2$ we get $$|d(p_n,q_n) - d(p_m,q_m)| < \epsilon$$ Thus, we have shown that $d(p_n,q_n)$ is Cauchy... in $\mathbb{R}$. Hence, by the completion of $\mathbb{R}$, $d(p_n,q_n)$ converges. QED. Is my attempt okay? $\endgroup$ – Curious Math Student Mar 15 '17 at 17:18
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    $\begingroup$ @CuriousMathStudent: Yes, but you shouldn't write $p_n \to p$ because it is not true (and you don't use it) and similarly for $q_n \to q$. Also, it should be $-\varepsilon$ in the second starred equation. $\endgroup$ – levap Mar 15 '17 at 19:01
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    $\begingroup$ Oh right! I forgot being Cauchy doesn't imply convergence unless we are in a complete metric space. Thank you for very much for your help! $\endgroup$ – Curious Math Student Mar 16 '17 at 2:08

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