TITLE: How to determine what kind of Cauchy sequences lie in a given space?
QUESTION [5 upvotes]: I understand the main principles of Cauchy sequences and metric spaces, but I have a particular question about determining whether or not a space is a complete space. If a space has all cauchy sequences converge to a point contained within that space, then it is a complete space. 
My question is: How do you determine if there is a cauchy sequence that doesn't converge?
I realize this is a rather vague question, so let me clarify a bit using an example, and you can just provide an answer to this particular example. For the space containing all real numbers $\mathbb R^1$, if our space is defined as the interval $(0,1)$ then it is not a complete space since there is a Cauchy sequence $x_n = \frac{1}{n}$ for which the limit converges to $0$, which is not contained in the space.

How did they come up with this sequence $x_n = \frac{1}{n}$? Is it just from experience working in the field of complete spaces and you start recognizing things like this? Or is there a more formal way to find it?
For that sequence, if $n \le 1 $ then $x_n$ is not in the given space. Do we need to specify that $n \gt 1$ when defining the sequence?

Thanks! If you can answer those two bullet points I think I'd understand this a bit more thoroughly =).

REPLY [2 votes]: This is more of a sharpening of what other answers have already said, but here goes:
Theorem: Let $(X,d)$ be a metric space, and let $Y \subset X$.  If $(Y,d)$ is complete, then $Y$ is closed in $X$.
Proof: If not, then there is $x \in X \setminus Y$ and a sequence $\{y_n\}$ in $Y$ with $y_n \rightarrow x$.  Since $\{y_n\}$ is convergent in $X$, it is Cauchy in $X$, hence it is Cauchy in $Y$.  Since limits in metric spaces are unique if they exist, $\{y_n\}$ is not convergent in $Y$.
This provides a nice approach to the present example, since $(0,1)$ is naturally given as a non-closed subset of the (complete) metric space $\mathbb{R}$.  Note also that $(0,1)$ is homeomorphic to $\mathbb{R}$, but this is not good enough: here it is really the "extrinsic" rather than the "intrinsic" topology that matters.
If we add an additional hypothesis on $X$, we can get a better result:
Theorem: If $(X,d)$ is a complete metric space and $Y \subset X$, then $(Y,d)$ is complete if and only if $Y$ is closed in $X$.
Proof: Even without the completeness hypothesis on $X$ we saw that it is necessary for $Y$ to be closed.  Conversely, if it is then every Cauchy sequence in $Y$ is convergent in the complete space $X$, and since $Y$ is closed, the limit is an element of $Y$.  So $Y$ is complete.
[Corollary: Any compact metric space is complete.
Sneaky proof: Let $X$ be compact, and embed it in its completion $\tilde{X}$.  Then $X$ is a compact subset of the Hausdorff space $\tilde{X}$, so it is closed.  Apply the theorem.]
Here is the moral, I think: if you are trying to decide whether a metric space is complete, embed it in a complete metric space and see whether the image is closed.  In theory this always succeeds since any metric space embeds in its completion.  In practice, the last sentence is a tautology but nevertheless the strategy seems to work in many cases.