Show that $X=\{0\} \cup \{\tfrac{1}{n}:n \in \mathbb{Z} - \{0\}\}$ is a complete metric space How could we show that the metric space
$$X=\{0\} \cup \{\frac{1}{n}:n \in \mathbb{Z} - \{0\}\}$$ with the metric it inherits as a subset of $\mathbb{R}$ is complete?
Thoughts
Complete metric spaces are those in which all Cauchy sequences converge to a point within the space. For any Cauchy sequence $(x_n)$ in the space, $|x_n|<1$ and so the sequence is bounded; bounded Cauchy sequences in $\mathbb{R}$ converge in $\mathbb{R}$ and so the limit, $x$ say, lies in $\mathbb{R}$.
Suppose that $x \notin X$. I imagine this leads to a contradiction but I can't see what it is.
Any help would be appreciated. Regards, MM.
 A: HINT (a variation of your approach): 
Show that $X$ is a closed subset of a complete metric space. [In fact, the same argument shows that every closed and bounded subset of the reals is complete.] 
A: Hint: If $\{x_n\}$ is any sequence in $X$, it has a  monotone subsequence. If this subsequence is not eventually constant, what must it  converge to?
A: I am not sure how much this will help, but it's an interesting approach nonetheless:


*

*Every compact metric space is complete (this is because every sequence must have a converging subsequence, and thus every Cauchy sequence is convergent).

*Every closed and bounded subset of $\mathbb R$ is a compact metric space.

*A set is closed if and only if it contains all its limit points.
A: One way to tackle the problem is to show that the given set is closed. You will have the required result as closedness implies completeness in complete metric spaces and $\mathbb{R}$ is complete. 
Observe that $X' = \{0\} $ Here $X'$ denotes the derived set of $X$ since the only limit point of $X$ is $0$. Given any other $1/n \in X $ choosing $ \delta = min \{|1/(n+1)-1/n|,|1/(n-1)-1/n| \} $ you can see that $(B(1/n, \delta)\setminus\{1/n\})\cap X = \emptyset. $ Thus $1/n \notin X'$ for all $n.$ Since $X' \subset X$ we have that $X$ is closed and hence it is complete.
If you want to tackle the question using the definition of a complete metric space, you must first observe that every Cauchy sequence in $X$ either becomes eventually constant or converges to $0$. 
Obviously, it is easier to approach the problem using the previous method.
A: The easiest approach is probably to show that $X$ is closed and use the fact that a closed subset of a complete metric space (with the subspace metric) is complete.
To see that $X$ is closed, there are many different ways. Here's one useful result:
If $U=\bigcup_{i\in I}U_{i}$ is a union of closed sets $U_{i}\subset Y$, so that for any $x\in Y\setminus U$ there exists a neighbourhood $B_{x}\subset Y$ such that $B_{x}\cap U_{i}\neq \emptyset$ for only finite amount of indices $i\in I$, then U is a closed subset of $Y$ (note that the index set $I$ can be arbitrary!). 
In your case, write the space $X$ as a union of singletons (closed sets), and note that origin is the only point of reals which doesn't have the above property and that $0\in X$.
Or, simply show that $\mathbb{R}\setminus X$ is open by considering the two cases $|x|\leq 1$ and $|x|>1$ when $x\in \mathbb{R}\setminus X$.
