MHB Complete Metric Space: X, d | Analysis/Explanation

shen07
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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.$$(-\infty,0)$$
 
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shen07 said:
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.
 
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 $[0,1)$.
 
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Deveno said:
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. :)
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.
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