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Let $d$ be metric on $\Bbb N$ defined by $$d(m,n) = \begin{cases} 1+\frac{1}{m+n}, m \ne n; \\ 0, m = n. \end{cases}$$ Show that the metric space $(\Bbb N, d)$ is complete.

Attempt: Let $(x_n)$ be an arbitrary Cauchy sequence in $(\Bbb N, d)$. Then $d(x_m,x_n) \to 0$ as $m,n \to \infty$. Now, if $x_m \ne x_n$, then $d(x_m,x_n) = 1+\frac{1}{x_m+x_n}>1 \not\to 0$. Hence, we must have $x_m=x_n$ and so, $d(x_m,x_n)=0<\epsilon$ for any $\epsilon>0$, for all $m,n \ge N$, and for some positive integer $N$. Thus, $(x_n)$ is convergent in $(\Bbb N, d)$. Therefore, $(\Bbb N, d)$ is a complete metric space.

Does this works?

math404
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    The proof is correct, but it can be made more rigorous (and clear) by explicitly setting the $N$ s.t. $\forall m,n\ge N, d(x_m, x_n)<1$. – erpxyr2001 Jul 16 '23 at 02:51

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While it has, in some essence, the key realizations, I don't think this constitutes a complete proof.

Your goal is to show that, provided $(x_n)$ is Cauchy, that it converges.

What does it mean for $(x_n)$ to converge, though? It means - in this space - that there is some $x \in \mathbb{N}$ such that $d(x_n,x) \to 0$ as $n \to \infty$.

Showing that $d(x_m,x_n) \to 0$ doesn't help anything -- in fact, you're already assuming that at the start, so you've not shown anything. (Especially considering that there exist Cauchy sequences which are not convergent in some spaces.) You need to find a prospective limit of the sequence -- and then prove, indeed, that it is the limit.

You have made the key realization that, if at any point in the sequence, we have $x_m \ne x_n$, then $d(x_m,x_n) > 1$. Consequently, there has to be some integer $N$ beyond which this stops, in the sense that $d(x_m,x_n) = 0$ for $m,n \ge N$ (because the sequence is supposed Cauchy).

The questions remain, then:

  • Why can I immediately say that $d(x_m,x_n) = 0$ and not just $d(x_m,x_n) \le 1$? (Pull back to your definition of $d$.)
  • What does this imply for the sequence $(x_n)$ and its behavior? (Try playing with a few examples of sequences.)
  • Consequently, what does this imply the limit of $(x_n)$ should be? And what is the proof that this limit is, indeed, the limit?
PrincessEev
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