Cauchy sequence without limit in a complete space?

1. Nov 11, 2009

LumenPlacidum

I know I'm doing something wrong here, but I can't find my mistake.

1) R^2 is a complete metric space under the ordinary Euclidean metric.
2) Consider the circle of radius 2, centered at the origin in R^2.
3) Construct a sequence {x_n} as follows:
x_1 is at the apex of the circle (0,2).
x_2 is counterclockwise around the circle, a distance of 1 away from x_1.
x_3 is counterclockwise around the circle, a distance of 1/2 away from x_2.
x_4 is counterclockwise around the circle, a distance of 1/3 away from x_3.
In general, x_n is counterclockwise around the circle, a distance of 1/(n-1) from x_n-1.
4) The distance between the x_n and x_n+1 is 1/n, which gets small as n gets large; so, {x_n} is a Cauchy sequence in R^2.
5) However, the total sum distance as n gets large diverges, since this is the harmonic series. Or, as you construct more and more points in the sequence, they continue to travel around the circle counterclockwise an arbitrarily-large number of times.
6) The sequence has no limit, since the sequence never settles on any one point on the circle.

This is in direct contradiction with R^2 being a complete metric space.

2. Nov 11, 2009

boboYO

First of all, why the circle? Your argument 'works' for the real line as well. The flaw is, you are misunderstanding the definition of a cauchy sequence. I will copy and paste from wikipedia:

emphasis mine; you only considered consecutive pairs :)

3. Nov 11, 2009

LumenPlacidum

Wait, really? Haha, to do my proof, that's exactly what I was hoping for. Silly me, misremembering Cauchy sequences.

Also, I had chosen the circle because it's easier to see how the sequence isn't converging, not that it's converging to something that's outside the space. I realize that either case would invalidate the completeness if it were a Cauchy sequence.

Thanks, though. Not sure how I missed that.

4. Nov 11, 2009

letmeknow

I have a question about this. If first we ignore the part about the sums and just consider the points and their distance from the previous terms, do these never approach a particular value on the circle? I see that it is Cauchy in a complete space so it must converge, I thought. It does seem that the sequence is Cauchy since given any e, we can find an n in N such that for all m>n, |x_m - x_{m+1}| < e.

I suppose that was my thinking at first. My second thought is that since the distance difference is the harmonic sequence, for any 2 points after n we don't have that the distance between them is ever necessarily less than e.

So the sequence isn't Cauchy for the second reason? Thanks

5. Nov 11, 2009

LumenPlacidum

Your first paragraph is wrong because you have to be able to get all of the terms after some Nth term to ALL be in an arbitrarily small epsilon-neighborhood.

Your second thought is correct. If epsilon is smaller than 4 (since I chose a circle of radius 2), then no matter how high you pick N, there will be some values of n and m that will give you points essentially on opposite sides of the circle, making them greater than epsilon distance from one another, implying that the sequence isn't Cauchy after all.

6. Nov 11, 2009

letmeknow

That's a great point. Thanks.