lim$_{n→∞}d(x_n, a) \neq d($lim$_{n→∞}x_n, a)$ in Metric Space - Implications A Metric Space $<M,d>$ is given by the Metric $M$ and distance function $d$
If there exists a Cauchy Sequence $x_n$ such that:
lim$_{n→∞}d(x_n, a) \neq d($lim$_{n→∞}x_n, a)$, for some $a \in M$ what can we imply about the Metric Space?
In particular, I'm curious if this affects the compactness/completeness of the metric space.
 A: In this case,the Cauchy sequence is not converging. The map $x\mapsto d(x,a)$ is uniformly continous, so $n\mapsto d(x_n, a)$ must converge.  The space cannot be complete and therefore it is not compact.
A: There's a misconception here that it would be good to clear up. Let's talk about something more concrete for a second. The limit of $\sin(x)$ as $x$ tends to infinity does not exist. So it is wrong to say $$\lim_{x\to\infty}\sin(x)=0.$$ But it is also wrong to say $$\lim_{x\to\infty}\sin(x)\ne0.$$There simply is no such thing as $\lim_{x\to\infty}\sin(x)$; if you say $\lim_{x\to\infty}\sin(x)\ne0$ that implies that there is such a thing as that limit, it just doesn't equal $0$.
So now for your question. Let's give that inequation a name. You ask what happens if $$\lim_{n\to\infty}d(x_n,a)\ne d(\lim_{n\to\infty}x_n,a).\quad(*)$$The answer is that (*) simply never happens. It's like asking what happens when $0=1$.
Because if $\lim x_n$ exists then we can prove that $(*)$ is false. But if $\lim x_n$ does not exist that does not mean $(*)$ is true. If $\lim x_n$ does not exist then $(*)$ is neither true nor false - it's meaingless, since it says something about non-existent things.
Why it matters: One might say this is not all that important; if someone cluefull says $\lim_{x\to\infty}\sin(x)\ne0$ that may not be quite correct, but a cluefull reader knows what he meant, so what's the harm?
The harm is when newbies talk about limits that don't exist. This all started with a certain exercise the OP had to do. He's said he had a solution. I've suspected his solution was wrong, and he just confirmed that what he did is indeed the wrong thing I suspected. And the wrong solution comes about precisely from talking about nonexistent limits as though they existed. Here it is:
Exercise Say $M=\{x\in\Bbb R^2:||x||<1\}$. Define a metric on  $M$ by saying $d(x,y)=2-||x||-||y||$ for $x\ne y$, $d(x,x)=0$. Show that $M$ is not  complete.
Wrong Solution Let $x_n=(1-1/n,0)$. Then $(x_n)$ is a Cauchy sequence in $M$. (Fine so far.) Now $\lim_{n\to\infty}x_n=(1,0)\notin M$. So $M$ is not complete. QED.
That solution really is wrong, not just poorly expressed or lacking details. If one somehow has the idea that we can always write $\lim x_n$ then it's easy to see where the wrong solution comes from; after all, what else could the limit be?
For anyone who thinks that "totally wrong" is putting it too strongly: if there is a sort of sensible version of the proof it would involve something like saying $S=\{x:||x||\le1\}$, and using the same formula to define a metric on $S$. But that same formula does not define a metric on $S$.
Or to put it another way: For that sequence $x_n$, any moderately plausible (but of course wrong) argument showing that $\lim x_n=(1,0)$ will also show that $\lim x_n=(0,1)$. (Because if we ignore the fact that the following $d$'s are undefined and just blindly appply the formula we get $d((1,0),x_n)=d((0,1),x_n)$.)
