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Let $S$ and $W$ be subsets of $\mathbb{R}$, with the usual metric,

\begin{align*} S &= \left\{\frac{1}{n} :n\in \mathbb{N}\right\}\cup\{0\} \\ W &= \left\{n+\frac{1}{n}: n\in\mathbb{N}\right\} \end{align*}

I have to check for completeness of these two subsets.

Here is my attempt:

$S$ being closed subset of $\mathbb{R}$ is complete.

$W$ is not closed in $\mathbb{R}$ since it doesn't contain all of its points and hence not complete.

I am not sure whether I am correct or not? Please help me with this.

Thanks

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@RossMillikan It was same as I have written. –  srijan Jun 3 '12 at 6:36
    
Then see my comment to the answer –  Ross Millikan Jun 3 '12 at 6:37
    
@RossMillikan What would be conclusion then? –  srijan Jun 3 '12 at 6:40
    
If your definition of complete is that all Cauchy sequences converge, it is complete. All Cauchy sequences in $W$ must be eventually constant and converge to something in $W$ –  Ross Millikan Jun 3 '12 at 6:42
    
@RossMillikan Do you meaan $W$ has discrete metric induced on it? I am confused with this. I am not able to visualize that $W$ looks like $\mathbb{N}$. –  srijan Jun 3 '12 at 6:47

1 Answer 1

up vote 4 down vote accepted

Not only is $S$ closed, it’s also bounded and therefore compact, so it’s certainly complete.

Correction: I somehow misread the definition of $W$ as $\Bbb Z^+\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$, so what I originally said about it is nonsense.

$$W=\left\{n+\frac1n:n\in\Bbb Z^+\right\}=\left\{2,\frac52,\frac{10}3,\dots,\frac{n^2+1}n\dots\right\}\;,$$

which is a closed, discrete set and therefore complete. The minimum distance between any two distinct points of $W$ is $1/2$, between $2$ and $5/2$. To prove this, suppose that $m,n\in\Bbb Z^+$ with $m<n$. Then since $\dfrac1m\le 1$, $m+\dfrac1m\le n<n+\dfrac1n$, and

$$\begin{align*} \left|\left(m+\frac1m\right)-\left(n+\frac1n\right)\right|&=n+\frac1n-\left(m+\frac1m\right)\\ &=n-m-\left(\frac1m-\frac1n\right)\\ &=n-m-\frac{n-m}{mn}\;. \end{align*}$$

Now $n-m<n$, and $mn\ge n$, so $\dfrac{n-m}{mn}<1$, and $n-m-\dfrac{n-m}{mn}>n-m-1$. This is at least $1$ unless $n=m+1$.

If $n=m+1$, $n-m-\dfrac{n-m}{mn}=1-\dfrac1{m(m+1)}$ is maximized when $m=1$, in which case it’s $\dfrac12$.

In all cases, therefore,

$$\left|\left(m+\frac1m\right)-\left(n+\frac1n\right)\right|\ge\frac12$$ when $1\le m<n$.

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More informative for me. Thank you very much sir. –  srijan Jun 3 '12 at 6:27
    
As $W$ is defined, $\frac 12$ is not an element of it. All the points of $W$ are isolated and $W$ looks a lot like 4\mathbb N$. –  Ross Millikan Jun 3 '12 at 6:46
    
If $n,m >2$, with $n\neq m$ then $|n-m+\frac{1}{n}-\frac{1}{m}| \geq \frac{1}{3}$. –  copper.hat Jun 3 '12 at 6:46
    
@copper.hat your answer shows that $W$ is not complete? –  srijan Jun 3 '12 at 6:49
1  
It shows that $W$ is closed. (My reasoning is not complete, as such, to finish I would need to show that all points are separated by some distance $\geq \lambda >0$.) –  copper.hat Jun 3 '12 at 6:56

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