Closed, bounded but not compact

[bender2]

Homework Statement

let |e^-x-e^-y| be a metric, x,y over R.
let X=[0,infinity) be a metric space.
prove that X is closed, bounded but not compact.

Homework Equations

The Attempt at a Solution

there is no problem for me to show that X is closed and bounded. but how do I prove it's not compact?
I assume it must be done with the use of Cauchy sequence. if x_n is Cauchy but it's not convergent then X is not complete and then it's not compact. but how do I right it down in algebraical form?

thanks in advance.

 

[LCKurtz]

What about trying to find a sequence of points each of which has its own open cover so that no finite subcover exists?

 

[bender2]

i'm afraid i don't know how to do it. can you show me please?

and is the way with Cauchy sequence correct?

 

[hunt_mat]

You can use the form of compactness with regard to subsequences. For every sequence x_{n} in the metric space then there is a convergent subsequence.

 

[HallsofIvy]

Normally one would prove that $[0, \infty)$, with the usual metric, is not compact, directly from the definition, "every open cover contains a finite subcover", by looking at the "open cover" $U_n= [0, n)$. Are those sets open in this metric?

 

[LCKurtz]

I think Halls means [0,n).

 

[HallsofIvy]

Yes, I did. Thanks. I will edit it.