A problem of checking completeness of some subsets of $\mathbb{R}$ 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 
 A: 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 $1\le m<n$.
