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The sequence ($\frac{1}{n}$) is a Cauchy sequence in $N$ but does not converge in $N$ as it converges to $0$.Then $N$ should not be complete. But $N$ is the closed subset of $R$ which is a complete metric space .Hence, $N$ should be complete and from here also link text Please clear my doubt.

And thanks for help in advance.

Sejy
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    $1/n$ is not a sequence in $\mathbb{N}$, it is a sequence in, say, $\mathbb{Q}$ or $\mathbb{R}$. – Ian Dec 01 '18 at 20:01
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    "The sequence (1/n) is a Cauchy sequence in N" No, it isn't. $\frac 1n$ is not a natural number. So ${\frac 1n} \not \subset \mathbb N$ and it is not a "sequence in $\mathbb N$". – fleablood Dec 01 '18 at 20:07

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$\frac 1n$ is not a Cauchy sequence in $\mathbb N$ as $\frac 1n \not \in \mathbb N$ for any $n \ne 1$.

To be a cauchy sequence OF NATURAL NUMBERS there must come a point where all the terms are less than $1$ apart. As all the terms ARE natural numbers[$*$] that means there comes a point where all the terms are equal. In other words the sequence is "constant for all but finite terms". And such sequences do converge to the constant.

More formally. If $\{m_i\} \subset \mathbb N$ is causchy so that for any $\epsilon > 0$ then there is an $M$ so that $n,p > M \implies |m_n - m_p| < \epsilon$. Then if $\epsilon < 1$ then there is some $M$ so that $n,p > M \implies |m_n - m_p | < 1$ which means $m_n = m_p$ for ALL $n,p> M$. And $m_n = m_p = c$ for some $c \in \mathbb N$.

SO $m_i\to c$. Yep. It's complete.

[$*$] That's what "being a sequence in $X$" means. It means all the terms are elements of $X$. So $\{\frac 1n\}$ must certainly is NOT a sequence "in $\mathbb N$".

fleablood
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