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In this thread, on the fourth question, is the following metric complete?

$4. \mathbb{R}, d(x,y)=|e^x-e^y|$

The example for it is not complete was: what about $\langle -n:n\in\Bbb N\rangle$?

My question is: since $d(x,y)=|e^x-e^y|$, it seems to me that the specific example definitely converges to $0$. Is it merely a typo?

Am I wrong? I feel $\langle n:n\in\Bbb N\rangle$ would alo suffice not to define a complete metric.

P.S I did put a comment there, but may it's an old thread so almost dead? You can just put a comment and I will delete this thread. to keep MSE clean.

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007resu
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1 Answers1

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This is a bit too long for a comment to the other thread.

If the sequence $\langle -n : n \in \mathbb{N} \rangle$ converges to $0$ in the $d$-metric, this means that for any $\epsilon > 0$ there is an $N$ such that $d ( -n , 0 ) < \epsilon$ for all $n \geq N$. However $$d ( -n , 0 ) = | e^{-n} - e^0 | = 1 - e^{-n},$$ and so as $n$ increases, so does the $d$-distance between $0$ and $-n$. Therefore $0$ cannot be the limit of the sequence in the $d$-metric. In a similar manner, you can show that for any real number $x$ there is an $N$ such that $d ( -(n+1) , x ) > d ( -n , x )$ for all $n \geq N$. Therefore $\langle -n : n \in \mathbb{N} \rangle$ cannot converge in the $d$-metric.

On the other hand the sequence $\langle n : n \in \mathbb{N} \rangle$ is not Cauchy in the $d$-metric, so it cannot converge. But the fact that it does not converge says nothing about the completeness of $d$.

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