Techniques to show a space is complete?

[pivoxa15]

Homework Statement

What are some ways to show that a space is complete?

It is tricky because to show completeness, must show every Cauchy sequence converges to a point in the space.

The Attempt at a Solution

One way is by contradiction? Suppose that there exists a Cauchy sequence that dosen't converge to a point in X than show it does converge?

 

[morphism]

This is too general. I guess a good thing to keep in mind is that a Cauchy sequence is convergent if it has a convergent subsequence. Sometimes this can be handy.

 

[pivoxa15]

How about I give a problem

Show that the metric d(x,y)=|x^3-y^3| is R complete. R is the real numbers.

 

[matt grime]

Show that this space and metric is equivalent to R under the usual metric, i.e. that the map sending x to x is a homeomorphism between (R,d) and (R,| |).

 

[pivoxa15]

Matt do you have an answer to the general question. The hard thing is 'every' cauchy sequence. Or are problems usually solved by recognising template complete metrics before hand and show homeomorphisms with the template ones.

For not complete, does it just invovle find a counter example?

 

[matt grime]

There will be no such thing as 'the method' - just show cauchy sequences have limits by whatever means you can. That's like askig - how do I show a function is continuous - depends on the function. How do I show a set is compact - depends on the set. I can't think of many places in maths where there is such a thing as 'the method' that always works.

 

[pivoxa15]

There will be no such thing as 'the method' - just show cauchy sequences have limits by whatever means you can. That's like askig - how do I show a function is continuous - depends on the function. How do I show a set is compact - depends on the set. I can't think of many places in maths where there is such a thing as 'the method' that always works.

What about the word 'every'? Is showing that one Cauchy sequence contains a limit in a metric enough? Or does the word every not matter as much? Because all Cauchy sequences are all the same in a certain metric. The key is the metric that determines whether a Cauchy sequence converges or not.

 

[morphism]

If it's sufficient for completeness that one Cauchy sequence converges, then completeness is a very useless definition, because in this sense every nonempty metric space is complete: just take any constant sequence - this is cauchy and convergent.

So you have to show that any Cauchy sequence converges.

 

[pivoxa15]

If it's sufficient for completeness that one Cauchy sequence converges, then completeness is a very useless definition, because in this sense every nonempty metric space is complete: just take any constant sequence - this is cauchy and convergent.

So you have to show that any Cauchy sequence converges.

Good point. Looks like one has to take some general, broader approaches. Considering what space the metric is in is a start and will give some indication to the solution of the problem.

 

[matt grime]

the empty metric space is also complete - since there are no sequences, the statement "for all sequences {x_n}, {x_n} cauchy implies {x_n} convergent" is vacuously true.

 

[morphism]

the empty metric space is also complete - since there are no sequences, the statement "for all sequences {x_n}, {x_n} cauchy implies {x_n} convergent" is vacuously true.

Yeah. The empty metric space is always complete.