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Suppose $\{x_n\}, \{y_n\}$ are Cauchy sequences in metric space $(X, d)$. Prove that the sequence $\{a_n\}$ defined by $a_n = d(x_n, y_n)$ converges in $\mathbb{R}$.

My Question. Before I write my proof attempt, I'll ask my question: I have the idea for this proof in mind, but I'm not so good at writing proofs using slick manipulations of inequalities. I came up with a manipulation that I want to use, but I'm not sure if I'm allowed to. So, my question is not necessarily about how to prove the above statement. (It has been proved on stack exchange here and here.) Rather my question is whether or not I'm allowed to manipulate inequalities in the way I have below, starting at the line "In symbols, we have..."

My Proof. Choose $\varepsilon >0$. Because $\{x_n\}$ is Cauchy, there exists $N \in \mathbb{N}$ such that $d(x_n, x_m) < \varepsilon/8$ for all $n, m \geq N$. A similar $M \in \mathbb{N}$ exists for sequence $\{y_n\}$. Choose $\max\{N, M\}$ and, without loss of generality, assume $\max\{N, M\} = N$. Then, for all $n \geq N$, we have $x_n \in B_{\varepsilon/4}(x_N)$ and $y_n \in B_{\varepsilon/4}(y_N)$. Let $$D = \{d(x,y) : x \in B_{\varepsilon/4}(x_N) \text{ and } y \in B_{\varepsilon/4}(y_N)\}.$$ Then for any $x_n, y_n$ with $n > N$, the distance $d(x_n, y_n)$ is bounded above by $\sup D = d(x_N, y_N) + \varepsilon/2$ and is bounded below by $\inf D = d(x_N, y_N) - \varepsilon/2$.

In symbols, we have

$$\begin{align} d(x_N, y_N) - \varepsilon/2 &< d(x_n, y_n) < d(x_N, y_N) + \varepsilon/2. \tag{1} \end{align}$$

Inequality (1) is still true for any other $m > N$, so substituting such an $m$ then multiplying through by $-1$, we obtain

$$\begin{align} -d(x_N, y_N) + \varepsilon/2 &> -d(x_m, y_m) > -d(x_N, y_N) - \varepsilon/2. \tag{2} \end{align}$$

Rearranging (2) so that its inequality signs face the same direction as (1), we can add (1) and (2) to obtain

$$\begin{align} -\varepsilon < d(x_n,y_n) - d(x_m, y_m) &< \varepsilon, \text{ i. e.} \\ |d(x_n, y_n) - d(x_m, y_m)| &< \varepsilon, \text{ meaning that} \\ |a_n - a_m| &< \varepsilon. \end{align}$$

Therefore, $\{a_n\}$ is Cauchy in $\mathbb{R}$, and so $\{a_n\}$ converges. (qed)

1Teaches2Learn
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  • From my perspective, your proof looks fine; I don't think your slickness causes an issue. There also may be something I don't see, as well. In terms of your question about inequalities, this MSE question may be helpful. https://math.stackexchange.com/questions/210620/subtraction-of-inequalities https://math.stackexchange.com/questions/210620/subtraction-of-inequalities –  Aug 15 '21 at 21:05
  • I have misgivings about your use of sup and inf, which you could resolve simply by not using those symbols. For example, $d(x_N, y_N) + \varepsilon/2$ is a legitimate upper bound and good enough for your purpose, whereas $\sup D$ is a least upper bound, which you have not shown and do not need. But the manipulations of the inequalities from $(1)$ to the end of the proof look perfectly fine. – David K Aug 15 '21 at 21:37
  • Thank you both as each of your comments is helpful. @DavidK I completely agree that the use of sup and inf is a bit sketchy as I've written it. I'll work on that part too. – 1Teaches2Learn Aug 16 '21 at 04:14
  • The bounds you have shown look fine to me, the only question is whether it is correct to call them sup and inf. No matter what you call them, they still are adequate bounds to prove what you need to prove. – David K Aug 16 '21 at 11:00

1 Answers1

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Yes, you can. So when you have $a<b$ and $c<d$ then $a-d < b-c$ holds true. Your proof looks quite fine, and there is no shady manipulation of inequalities happening.

Arturo Magidin
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Lazy
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