Is "completeness" a well defined notion for a pseudo-metric? Let $(X,d)$ be a pseudometric space,

Definition: any sequence $\{x_n\}$ in $X$ is $Cauchy$ if for every $\epsilon>0$ there is a $n_0$ such that for every $m,n>n_0$ it results $d(x_m,x_n)<\epsilon$.
Definition: a point $x\in X$ is a limit of $\{x_n\}$ if for every $\epsilon>0$ there is a $n_0$ such that for every $n>n_0$ it results $d(x,x_n)<\epsilon$.

Does it make sense to state that $(X,d)$ is $complete$ if every Cauchy sequence has a limit (although not unique)?
I would answer "yes": indeed if $C\subset X$ is such that for every $c_1,c_2\in C$ it results $d(c_1,c_2)=0$ then every sequence $\{x_n\}$ in $C$ is Cauchy, but it is also true that every point $C$ is a limit point of $\{x_n\}$. Of course, in a pseudometric space the limit has no reason to be unique, but this should not be a problem in the definition of a complete pseudometric space.
What do you think?
 A: The definition of a convergent sequence does not require that the limit is unique. Indeed, if the space is not Hausdorff, then there is no reason to expect that (e.g. consider any set with a dense point, and a constant sequence of that point).
Completeness states "Every Cauchy sequence is convergent". And there is nowhere in the definitions anything about uniqueness of the limit.
Now. Using the completeness, that's a whole other story, and if you need to be picky about which limit you choose, that might be something that you can't do off-hand when there are several options. But this has nothing to do with the definition of completeness.
The key point here is that any two limits of a sequence in a pseudo-metric space must have pseudo-distance of $0$.

Just for fun, here's a nice way to see why the definition makes sense.
Given a pseudo-metric space $(X,\rho)$, the metric reflection of $X$ is the quotient space given by the relation $x\sim y\iff\rho(x,y)=0$. Then there is a natural metric $d$ on $X/{\sim}$, given by $d([x],[y])=\rho(x,y)$.
It is not hard to verify that $d$ does not depend on the choice of representatives, and that it is indeed a metric.
Theorem. If $(X,\rho)$ is a pseudo-metric space, then it is complete if and only if its metric reflection is complete.
Proof. Suppose that $X$ is complete, and $[x_n]$ is a Cauchy sequence, then for every $\varepsilon>0$, there is some $n$ such that for all $m,k>n$, $d([x_m],[x_k])<\varepsilon$, so by definition $\rho(x_m,x_k)<\varepsilon$, and therefore $x_n$ makes a Cauchy sequence. Since it has a limit (possibly many limits) $x$, and it is easy to check that $[x]$ is indeed the limit of $[x_n]$ in $X/{\sim}$.
In the other direction, if $X/{\sim}$ is complete, and $x_n$ is a Cauchy sequence in $X$, then $[x_n]$ is Cauchy in $X/{\sim}$, so it has a limit $[x]$. And so $x$ serves as a limit in $X$. $\quad\square$
So one can define the completeness of pseudo-metric spaces by the completeness of their reflection.

Note, as remarked in the comments, that there is no unique completion of a pseudo-metric space, since we can always add arbitrarily many points as a limit of some Cauchy sequence. But one can take one which is "sufficiently canonical" that adds a single point to any missing limit.
