Is every closure of a metric space a completion? I know that every completion is a closure of a metric space, since every convergent sequence is cauchy and  and the limit of that sequence will exist within the completion. 
At the same time, from my understanding, every cauchy sequence will bunch closer together and get arbitrarily close to something, but it is just a question as to whether or not that element it gets closer to actually exists in the space.
This leads me to the question as to whether every closure of a metric space is a completion, because we would just be adding the limits to sequences which exist outside of the original space, including the limits of nonconvergent cauchy sequences. 
So is there an example of a closure which is not a completion? Or are these notions  equivalent?
 A: Closures and completions are very different beasts. The process of completion takes a metric space and spits out a new one, with some new limits added. So, the completion of $\mathbb Q$ under the usual metric is $\mathbb R$. No matter what.
Closure, on the other hand, takes a metric space and a subset thereof, and puts in any points from that metric space that are limits of the subset. It doesn't, however, put in any points which were "missing" from the space. So, if we're working in $\mathbb Q$, then the closure of the subset $(0,1)\cap \mathbb Q$ is $[0,1]\cap\mathbb Q$, which is still missing irrationals since the metric space we're working in doesn't have them. More dramatically, if $(M,d)$ is any metric space, then $M$ is closed, but might not be complete.
It is notable that, in complete metric spaces, the closure of a subset $S$ is essentially (i.e. isomorphic to) the completion of the sub-metric space represented by $S$.
A: Your question doesn't quite make sense. Every metric space is closed by definition. For example if you consider $(0,1)$ with the regular metric, the closure of $(0,1)$ is just $(0,1)$.
