About the definition of completion of a metric space $R$. The definition of completion of a metric space $R$ is the following:

Definition 1:
Given a metric space $R$ with closure $[R]$, a complete metric space $R^*$ is called a completion of $R$ if $R\subset R^*$ and $[R]=R^*$, i.e., if $R$ is a subset of $R^*$ everywhere dense in $R^*$.

Why is the following definition of mine bad?

My definition:
Given a metric space $R$, a complete metric space $R^*$ is called a completion of $R$ if $R\subset R^*$.


jjagmath, thank you very much for your comment.
Is the following definition equivalent to the definition 1?

My definition 2:
Given a metric space $R$, a complete metric space $R^*$ is called a completion of $R$ if $R^*$ is the smallest complete metric space which includes $R$.

 A: Take as our starting metric space $[0,1)$. Obviously the "right" completion is $[0,1]$, that is:

*

*take the original space;


*add a new point to it, namely $1$;


*extend the metric to set $d(a,1)=1-a$ for each $a\in [0,1)$.
However, this isn't the only thing we could do. For example, we could also look at the metric space whose underlying set is $[0,1)\cup\{17\}$ with distance function $f$ given by $f(x,y)=\vert x-y\vert$ if $x,y\in [0,1)$, $f(17,17)=0$, and for $a\in [0,1)$ we set $f(a,17)=f(17,a)=1-a$.
This sort of issue indicates why "definitions" like "the smallest complete metric space containing $R$" are going to be problematic: they require us to compare objects which don't, on the face of it, admit any sort of meaningful comparison. That said, there are a couple ways to make your proposed definition precise, which I'll phrase as theorems (using the standard definition of "completion"):

*

*Suppose $R\subseteq R^*$ are metric spaces with $R^*$ complete. Then $R^*$ is a completion of $R$ iff no proper subspace of $R^*$ containing $R$ is complete. (So completions are "internally minimal.")


*Suppose $R\subseteq R^*$ are metric spaces with $R^*$ complete. Then $R^*$ is a completion of $R$ iff for every complete metric space $S$ with $R\subseteq S$ there is a unique isometric embedding of $R^*$ into $S$ which is the identity on $R$. (So completions are "minimal with respect to comparisons via isometries.")
The second notion isn't as snappy, but it's actually getting at a deeper idea than the first (that of figuring out how to compare very disparate objects by looking for maps between them, especially unique maps, satisfying certain nice properties).
A: The charm of universal problems.-
Too long for a comment, I would like to highlight the point of view of universal problems which is evoked as second point in the end of Noah's answer.
Indeed, completions of a metric space are not unique (but they are isomorphic) and, oftentimes, in the constructions of the "working mathematician", the original space $R$ does not come naturally as a subspace of the completed one $R^*$. As example, one can cite the celebrated completion $\mathbb{R}$ of $\mathbb{Q}$ which can come from

*

* Limits of Cauchy sequences

* Order i.e. Dedekind cuts, see [1]

* Classes of functions, see Eudoxus real numbers [2] 
   
 The following definition, IMHO, is the best. 
Definition.- Let $(R,d)$  be a metric space. A completion of $R$ is a pair $(u,R^*)$, where
 
*

* $(R^*,d^*)$ is a complete metric space

* $u:\ R\to R^*$ is uniformly continuous   

Such that, for all  $w:\ R\to T$ uniformly continuous 
(where $(T,d_T)$ is a complete metric space), it exists a unique $v:\ R^*\to T$, uniformly continuous, with $w=v\circ u$. 
As for every such scheme (algebraic closure, tensor products, enveloping algebras etc.), one shows  easily that all completions are isomorphic (i.e. “completion” is unique “ up to isomorphism”).

At the end of the day, if one prefers inclusion maps rather than embeddings, one can even - at the cost of a bit of surgery - tailor solutions $(u,R^*)$ where $u$ is an inclusion map.
[1] Dedekind cuts
[2] Eudoxus real numbers
