Complete Metric Space: X, d | Analysis/Explanation

[shen07]

Hi i am confused of the following question.

Suppose we have a Metric Space (X,d), where d is the usual metric. Now are the following subsets complete, if so why??

1.$$X=[0,1]$$
2.$$X=[0,1)$$
3.$$X=[0,\infty)$$
4.$$(-\infty,0)$$

 

[Sudharaka]

Hi i am confused of the following question.

Suppose we have a Metric Space (X,d), where d is the usual metric. Now are the following subsets complete, if so why??

1.$$X=[0,1]$$
2.$$X=[0,1)$$
3.$$X=[0,\infty)$$
4.$$(-\infty,0)$$

Hi she07, :)

Since any closed subspace of a complete metric space (in this case the complete metric space is the set of real numbers with the usual metric) is complete, the fist interval is complete (refer >>this<<).

Conversely, a complete subset of a metric space is closed. Therefore the second third and fourth intervals are not complete since they are not closed.

 

[Deveno]

I think the third interval is indeed closed, since its complement:

$(-\infty,0)$ is open.To see why the second interval is NOT complete, consider the sequence:$a_n = \dfrac{n}{n+1}$.It is clear that the set $A = \{a_n: n \in \Bbb N\} \subseteq [0,1)$ and that $\{1\}$ is a limit point of $A$ (and thus of $[0,1)$), but is not in $[0,1)$.Put another way, the sequence above is Cauchy, but not convergent to a point in $[0,1)$.

 

[Sudharaka]

I think the third interval is indeed closed, since its complement:

$(-\infty,0)$ is open.To see why the second interval is NOT complete, consider the sequence:$a_n = \dfrac{n}{n+1}$.It is clear that the set $A = \{a_n: n \in \Bbb N\} \subseteq [0,1)$ and that $\{1\}$ is a limit point of $A$ (and thus of $[0,1)$), but is not in $[0,1)$.

Put another way, the sequence above is Cauchy, but not convergent to a point in $[0,1)$.

Thanks for pointing out the mistake. Yep, the second interval is indeed closed. :)