# Cauchy Sequence in Metric Space

1. Feb 20, 2010

### BSCowboy

For a metric space (X,d), prove that a Cauchy sequence {xn} has the property d(xn-xn+1)--->0 as n--->\infty

In working this proof, is it really as simple as letting m=n+1?

2. Feb 21, 2010

### HallsofIvy

Staff Emeritus
Yes, it really is that easy! The definition of "Cauchy sequence" is that $d(x_n, x_m)$ goes to 0 as m and n go to infinity. Since that is true, in particular, $d(x_n, x_{n+1})$ must go to 0 as n goes to infinity.

Notice, however, that the converse is not true. If $d(x_n, x_{n+1})$ goes to 0, it does NOT follow that $d(x_n,x_m)$ goes to 0.

3. Feb 21, 2010

### BSCowboy

Thanks for the input.

An example of a sequence in which $$d(x_n,x_{n+1})\rightarrow 0$$ , but $$d(x_n,x_m)\not\rightarrow 0$$ would be let $$x_n=ln(n)$$

Last edited: Feb 21, 2010
4. Feb 21, 2010

### HallsofIvy

Staff Emeritus
I'll have to think about that! I was thinking of
$$a_n= \sum{i= 1}^n \frac{1}{i}= \frac{1}{n}+ \frac{2}{n}+ \frac{3}{n}+ \cdot\cdot\cdot+ \frac{1}{n}$$

Then $|a_n- a_{n-1}= \frac{1}{n}$ goes to 0 as n goes to infinity but the series is not Cauchy since it is well known that the harmonic series does not converge.

(After about 30 seconds of thought I see that ln(n)- ln(n+1)= ln(n/(n+1)) so, in fact, your seires is a variation of mine.