# General approach to determine completeness of metric space

I've looked at a few questions online asking to determine the completeness of Metric Spaces.

2 such examples of metric spaces $(M,d)$:

1) $M = \{ (x,y) \in \mathbb{R}^2 \space : y>0$ or $x=0=y \}$

$\space \space$ $d((x,y),(a,b)) = min \{ max \{ |x-a|,|y-b| \},y+b \}$

$\space$

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

$\space \space$ $d(x,y) = \left\{\begin{matrix} 0 \iff x=y\\ 2-||x||-||y|| \iff x \neq y \end{matrix}\right.$

$\space$

I'm less interested in the answer and more interested in the general approach I might take to determine the completeness of metric spaces (not just the 2 above).

People seem to know an appropriate Cauchy Sequence to use as a counter example, or the correct proofs off the top of their heads. What might their thought process be?

I'm very new to metric spaces so I'd be very very grateful if I was given a more intuitive explanation rather than some formal maths.

Thanks!

Both of this examples are rather counter-intuitive but let's look at the second: it seems that points $x,y$ "close to $1$ in the usual metric" (such as $0.999$ and $0.9999$) are also very close to each other in your metric. So one would expect that if the space were complete, the sequence $0.9,0.99,0.999,\ldots$ would converge to something, as the element are very close to each other. But you can easily show that this sequence has no limit (intuitively, the limit should be $1$---even in your strange metric---but $1$ is missing in the space). To finish the proof, of course, you need to formalize these notions which should be easy.
So one general key for proving incompleteness is to try to find a sequence of numbers that are very close to each other (this is formalized in the notion of Cauchy sequence) but have no limit. One example I recommend to have in mind is the incompleteness of $\Bbb Q$, where a sequence of rationals "converging to $\sqrt{2}$" has no limit in $\Bbb Q$. In other words, something is missing there. In the second of your examples, $1$ "is missing". So the right question to start with is
To prove completeness, such as in your first example, may be harder: in this case, I would probably split the two cases: either a Cauchy sequence has $y$-coordinates bounded from below by a positive number and then your metric is, in a neighborhood of your sequence, locally the usual max-metric on the closed upper half-plane (I leave to you to work out the details) which is complete, or the $y$-coordinates of your sequence approach to zero and then the limit is clearly $(0,0)$. But the intuition is the same: if the $y$-coordinates are very small, then the points are "close to $(0,0)$", an element of our space.
• Nice answer! I'm wondering if I could use the following approach for cases where the space is a subset of $\mathbb{R}^n$: Prove the space is compact by showing it is closed and bounded. This would then imply it is complete. It seems like this might be an easy approach for some specific spaces? – Gregory Peck Jan 3 '16 at 13:41
• @GregoryPeck Every compact metric space is complete. However, be careful: if it is a subset of $\Bbb R^n$ endowed with some strange metric, then it may be hard to check boundedness and closedness means nothing (closed in what?). Further, bounded and closed doesn't imply compact in general metric spaces, not even in complete metric spaces. math.stackexchange.com/questions/627667/… – Peter Franek Jan 3 '16 at 14:23
• @GregoryPeck I tried to visualize it. Everywhere except near the $x$-axis, it behave "normally" like $\Bbb R^n$. So the only counterexamples that came to my mind were such that $y$ converges to $0$. All the part of the space with small $y$ coordinate is "very close" to $(0,0)$. But if $y$-coordinates converge to $0$, then, you see, everything is ok. At least I hope I'm not missing something tricky :) – Peter Franek Jan 3 '16 at 14:24