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According to Wikipedia:

A metric space $M$ is said to be complete if every Cauchy sequence converges in $M$

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A metric space $M$ is compact if every sequence in $M$ has a subsequence that converges to a point in $M$

I can't seem to find a situation where a complete metric space is not compact or vice versa.

First of all, why can't we say $M$ is complete if every sequence converges in $M$. Since if a sequence converges to a point outside $M$ it is clearly not complete?

And so if every sequence converges in $M$, then clearly every sequence has a subsequence which converges in $M$ and hence it is also compact (if it is complete).

And if every sequence has a subsequence which converges to $M$ then doesn't the sequence itself converge to $M$ in which case the compact space is also complete?

Apologies if I'm totally off track.

If someone could provide me with some examples which show the difference between the two I'd be very grateful. Preferably an example that gives me a much better intuitive understanding, because I think my main problem is intuition, I go a bit mad trying to understand rigorous definitions.

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    $\begingroup$ A metric space is compact if and only if it is complete and totally bounded. $\endgroup$ – Watson Dec 30 '15 at 17:57
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The real line $\mathbb{R}$ is complete but not compact. The key word in the definition of completeness is "Cauchy". Note that the definition of compactness does not speak of Cauchy sequences but rather of arbitrary sequences.

Here $\mathbb{R}$ is not compact because the sequence $u_n=n$ does not contain a convergent subsequence.

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  • $\begingroup$ Of course, thanks! $\endgroup$ – Gregory Peck Dec 30 '15 at 18:54
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Everything is explained once I answer this question, from which all other questions follow:

First of all, why can't we say $M$ is complete if every sequence converges in $M$. Since if a sequence converges to a point outside $M$ it is clearly not complete?

By your proposed definition, $\mathbb{R}$ would not be complete, since it is easy to furnish divergent sequences ($(-1)^n$, for instance). What you want to say is "OK, but every sequence that should converge must con... oh wait, that is the definition of completeness".

For an example of a complete but not compact space, $\mathbb{R}$ suffices.

Compactness implies completeness. To see that is easy. Take a Cauchy sequence. Since we are on a compact set, it has a convergent subsequence. But a Cauchy sequence with a convergent subsequence must converge (this is a good exercise, if you don't know this fact).

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  • $\begingroup$ This is also a great answer! Cheers! $\endgroup$ – Gregory Peck Dec 30 '15 at 18:54

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