Is (0,\infty) a Complete Metric Space?

[AxiomOfChoice]

I am correct in asserting that $(0,\infty)$ is NOT complete, right? It seems the sequence $x_n = 1/n$ is a Cauchy sequence that does not converge in this metric space (in the standard Euclidean metric), since 0 is not in the space.

Also, doesn't it follow from the completeness of $\mathbb R$ that it's NOT possible to construct an unbounded Cauchy sequence (i.e., a sequence of real numbers that converges to $\infty$ but whose terms get closer and closer together)? Is there a way to prove this without appealing to the fact that the reals are complete?

 

[mathman]

Let x_n=Σ(k=1,n) 1/k. |x_n-x_n+1|-> 0.
However |x_n-x_m| may -> ∞ as |n-m| -> ∞.

So you need to be precise as to what you are referring to. If it is to the usual idea of a Cauchy sequence (n and m independent) then it is not possible, since fixing n and letting m increase will lead to the second part of the example.

 

[Landau]

Also, doesn't it follow from the completeness of $\mathbb R$ that it's NOT possible to construct an unbounded Cauchy sequence (i.e., a sequence of real numbers that converges to $\infty$ but whose terms get closer and closer together)?

Of course it does. Since R is complete, Cauchy sequences and convergent sequences are the same thing. An unbounded sequence is not convergent, hence not Cauchy.

Is there a way to prove this without appealing to the fact that the reals are complete?

I'm not following you. You want to prove that completeness of R implies something, without using the completeness of R?

 

[AxiomOfChoice]

Of course it does. Since R is complete, Cauchy sequences and convergent sequences are the same thing. An unbounded sequence is not convergent, hence not Cauchy.

Ok, that's nice. Thanks.

I'm not following you. You want to prove that completeness of R implies something, without using the completeness of R?

Certainly not! I'm asking for a proof of the fact that if $\{x_n\}$ is Cauchy, then $\{x_n\}$ cannot trail off to infinity, without simply invoking the completeness of $\mathbb R$ (in which case it's trivial, as you pointed out above)...is there a proof that proceeds directly from the definition of Cauchy?

 

[Landau]

I'm asking for a proof of the fact that if $\{x_n\}$ is Cauchy, then $\{x_n\}$ cannot trail off to infinity, without simply invoking the completeness of $\mathbb R$ (in which case it's trivial, as you pointed out above)...is there a proof that proceeds directly from the definition of Cauchy?

Yes, see mathman's last remark. Here is a formal proof:

Suppose (x_n)_n is unbounded. Let's prove that it is not Cauchy: for all N there exists m,n>N such that |x_n-x_m|>1 (this is the negation of the definition of Cauchy sequence, with epsilon=1).

Let N be arbitrary. Take some m>N. Now take n large enough so that |x_n|>|x_m|+1 (possible by unboundedness). Then

$$|x_n-x_m|\geq||x_n|-|x_m||>1.$$

 

[adriank]

A subspace of a complete metric space is complete if and only if it is closed.