Does lim$_{a \rightarrow b } \space d(a,b) = 0 $ imply completeness in a metric space?

Question

Suppose $<M,d>$ is a metric space.

Does lim$_{a \rightarrow b } \space d(a,b) = 0 $ imply completeness in a metric space?

Or maybe lim$_{a \rightarrow b } \space d(a,b) \neq 0 $ implies incompleteness in a metric space?

It seems to have been correct with 3 examples I've tried it on.

But intuitively, it doesn't seem satisfactory to me.

It seems to show the sequence is Cauchy but I can't understand why this might imply convergence. (A friend said it did).

Any ideas?

Suppose we have the incomplete metric space$ (M, d) $ s.t.

$M=\{ x \in \mathbb{R}^2 :||x|| < 1 \} $

$d(x, y)= 2 - ||x||-||y|| $ if $x \neq y$, and $0$ otherwise

The original question I asked is only true here when $||b||=1$ which can't be the case given the definition of the space $M$

Answer 1

The statement $$\lim_{a\rightarrow b}d(a,b)=0$$ is true of all metric spaces. Just expand what it means:

For all $\varepsilon$, there is some $\delta$ such that if $d(a,b)<\delta$ then $d(a,b)<\varepsilon$.

Which is trivially satisfied by putting $\delta=\varepsilon$. So, in particular, it tells you nothing about completeness. In words, the statement says (more or less):

All sequences converging to $b$ get arbitrarily close to $b$.

which is very different from the notion of completeness, since completeness starts with a Cauchy sequence and says it converges, whereas here we've started with a sequence we already knew converged and said it converged.