Is a Cauchy Sequence in a Metric Space Characterized by d(xn, xn+1) → 0?

[BSCowboy]

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

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

 

[HallsofIvy]

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.

 

[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)

 

[HallsofIvy]

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.