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. :)

 

[blue_raver22]

A complete metric space is a metric space in which every Cauchy sequence converges. In other words, a complete metric space is one in which there are no "missing points" or "gaps" between the elements. In this context, a Cauchy sequence is a sequence of points in the metric space such that for any given distance, there exists a point in the sequence after which all other points are within that distance.

1. The subset X=[0,1] is a closed and bounded subset of the real numbers. Since the real numbers are a complete metric space, any subset that is closed and bounded will also be complete. Therefore, X=[0,1] is a complete metric space.

2. The subset X=[0,1) is an open and bounded subset of the real numbers. However, the endpoint 1 is not included in the subset. This means that there is a "gap" between the elements in the subset and the endpoint 1. Therefore, X=[0,1) is not a complete metric space.

3. The subset X=[0,∞) is a closed and unbounded subset of the real numbers. Since the real numbers are a complete metric space, any closed subset will also be complete. However, since X is unbounded, there may be Cauchy sequences that do not converge in X. Therefore, X=[0,∞) is a complete metric space, but there may be Cauchy sequences that do not converge in X.

4. The subset X=(-∞,0) is an open and unbounded subset of the real numbers. Similarly to X=[0,1), there is a "gap" between the elements in the subset and the endpoint -∞. Therefore, X=(-∞,0) is not a complete metric space.

In summary, subsets that are closed and bounded or closed and unbounded in a complete metric space will also be complete. However, subsets that are open and bounded or open and unbounded may not be complete due to the potential for "gaps" between the elements.