Equivalence classes and metric spaces Let $(X,d)$ be a metric space and let $S$ be the set of Cauchy sequences in $S$. Define "~" in $S$ by "$\{s_k\}$ ~ $\{t_k\}$" to mean that $d(s_k,t_k) \to 0$ as $k \to \infty$. Let $X'$ denote the set of equivalence classes of $S$ and let $s'$ denote the equivalence class of $s = \{s_k\}$. 
I know $\phi(s',t') = \lim\limits_{k \to \infty} d(s_k,t_k)$ is a metric on $X'$. 
Show $(X',\phi)$ is complete.
For $x \in X$, define $x'$ to be the equivalence class of the constant sequence ${x,x,...}$. Show that the function $x \to x'$ is an isometry of $X$ onto a dense subset of $X$. 
If anyone can share a proof, or steer me in the right direction, maybe tell me what it is that I need to prove specifically, that would be great.
 A: I don't think you've defined $\phi$, but it ought to be something like $\phi(\{x_k\},\{y_\ell\}) := \lim_{k\to\infty} d(x_k,y_k)$. 
I think firstly you need to check that this is well-defined and a metric on $X'$ (i.e. for $a,b,c \in X'$, $\phi(a,b) = \phi(b,a)$, $\phi(a,b) = 0 \implies a = b$, and the triangle inequality $\phi(a,b) \le \phi(a,c) + \phi(c,b)$)
Completeness means "Cauchy sequences converge", i.e. if $\{a_n\}$ is a sequence in $X'$ (note: a sequence of equivalence classes of sequences!) that is Cauchy, then there exists $a \in X'$ such that $\lim_{n\to\infty} \phi(a_n,a) = 0$ (I think something like the "diagonal" sequence $\{a_{n,n}\}$, that is the sequence with $n$th term equal to the $n$th of $a_n$, should work).
To show that the map $x \mapsto x'$ is an isometry (I would prefer to say an isometric embedding) means we need to check that it satisfies $\phi(x',y') = d(x,y)$. If my definition of $\phi$ is right this should be a one-line proof.
Finally we want to show that the image of the map is dense, that is, given any element $x \in X'$ and any $\epsilon > 0$, there exists a constant sequence $y' = y,y,\dots$ such that $\phi(x,y') < \epsilon$. This should follow pretty straightforwardly from the fact that $x$ (as a sequence in $X$) is Cauchy.
