Cauchy sequence without limit in a complete space?

[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 anyone point on the circle.

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

 

[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:

In other words, suppose a pre-assigned positive real value e is chosen. However small e is, starting from a Cauchy sequence and eliminating terms one by one from the start, after a finite number of steps, any pair chosen from the remaining terms will be within distance e of each other.

emphasis mine; you only considered consecutive pairs :)

 

[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.

 

[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

 

[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.

 

[letmeknow]

That's a great point. Thanks.