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Prove $(\mathbb{N},d_2)$ is a complete metric space.

Attempt: So I need to show that every Cauchy sequence in this metric space converges. Presumably all of these convergent Cauchy sequences would be eventually constant -- otherwise they wouldn't converge in $(\mathbb{N},d_2)$.

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    Just a side-question, I'm not familiar with this: what is $d_2$ here?2012-02-29
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    the usual metric.2012-02-29
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    Hello @Emir, I was pretty sure that it meant a metric, regarding the context, but not so sure of why there was an index 2. I was asking out of pure curiosity and hoping you could give me a short definition of the 'usual metric'. Is it just $d_2(m,n):=|m-n|$? If so, your attempt is the right way. Nevermind, Brian M. Scott already formalized the argument a bit ;)2012-02-29
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    I have downvoted this because you assume everyone knows what $d_2$ means in your post. You should include definitions the next time. In the body of the question, not in later comments.2012-02-29
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    @Rand,@Asaf: $d_2$ is a fairly common notation for the Euclidean metric. It’s probably worth asking for confirmation, but I can understand why Emir wouldn’t have thought it necessary to mention.2012-02-29

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