Every closed ball in X is complete, then X is complete

In summary: A 'ball' in a metric space is just the set of all points x in the space such that |d(x,x0)|<=R. It doesn't have to be spherical.
  • #1
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Homework Statement .
Prove that if every closed ball in a metric space X is a complete subspace, then X is complete.


The attempt at a solution.

Let ##\{x_n\}## be a cauchy sequence in X. Then, for ##ε=1##, ##\exists## ##n_0 \in \mathbb N##: ##\forall## ##m≥n_0##,##n≥n_0## ##d(x_m,x_n)<1##. Consider the closed ball centered at ##x_{n_0}## of radius 1. Then, the subsequence ##\{x'_n\}## such that ##x'_n=x_{n_0 + n-1}## is a subsequence of ##\{x_n\}## included in ##K(x_{n_0},1)## . By hypothesis, the ball is a complete subspace and the subsequence ##\{x'_n\}## is cauchy. But then ##\{x'_n\}## converges to some point ##x \in K(x_{n_0},1)## This implies ##\{x'_n\}## converges to the same point x in the metric space X. So the original sequence ##\{x_n\}## has a convergent subsequence that converges to the point x, this means that ##\{x_n\}## converges to x. I've proved that an arbitrary cauchy sequence in X is convergent, this implies every cauchy sequence in X is convergent, then X is a complete metric space.


Is my proof correct?
 
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  • #2
mahler1 said:
Homework Statement .
Prove that if every closed ball in a metric space X is a complete subspace, then X is complete.


The attempt at a solution.

Let ##\{x_n\}## be a cauchy sequence in X. Then, for ##ε=1##, ##\exists## ##n_0 \in \mathbb N##: ##\forall## ##m≥n_0##,##n≥n_0## ##d(x_m,x_n)<1##. Consider the closed ball centered at ##x_{n_0}## of radius 1. Then, the subsequence ##\{x'_n\}## such that ##x'_n=x_{n_0 + n-1}## is a subsequence of ##\{x_n\}## included in ##K(x_{n_0},1)## . By hypothesis, the ball is a complete subspace and the subsequence ##\{x'_n\}## is cauchy. But then ##\{x'_n\}## converges to some point ##x \in K(x_{n_0},1)## This implies ##\{x'_n\}## converges to the same point x in the metric space X. So the original sequence ##\{x_n\}## has a convergent subsequence that converges to the point x, this means that ##\{x_n\}## converges to x. I've proved that an arbitrary cauchy sequence in X is convergent, this implies every cauchy sequence in X is convergent, then X is a complete metric space.


Is my proof correct?

Why are you having doubts? Eventually the tail of every cauchy sequence is contained in a closed ball. The closed ball is complete. So the sequence is convergent.
 
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  • #3
Dick said:
Why are you having doubts? Eventually the tail of every cauchy sequence is contained in a closed ball. The closed ball is complete. So the sequence is convergent.

Thank you, I'm new to proofs and mathematics in general so I usually have strong doubts on whether I am doing the exercises right or not. I suppose it's a newbie's insecurity.
 
  • #4
mahler1 said:
Thank you, I'm new to proofs and mathematics in general so I usually have strong doubts on whether I am doing the exercises right or not. I suppose it's a newbie's insecurity.

Well, I thought that was a pretty fine and clear proof. I think you should start feeling less insecure.
 
  • #5
The only thing is that you don't want to say ε=1.

Because your metric space may be "smaller" than the unit closed ball.

You could just say for some ε. But the proof seems fine
 
  • #6
Jufro said:
The only thing is that you don't want to say ε=1.

Because your metric space may be "smaller" than the unit closed ball.

You could just say for some ε. But the proof seems fine

A 'ball' in a metric space is just the set of all points x in the space such that |d(x,x0)|<=R. It doesn't have to be spherical. The proof really doesn't need any qualification.
 

What does it mean for a closed ball in X to be complete?

A closed ball in X is complete if every Cauchy sequence within the ball converges to a point within the ball. This means that there are no "missing" points within the ball and the sequence of points within the ball does not "escape" to points outside of the ball.

How does the completeness of closed balls relate to the completeness of X?

If every closed ball in X is complete, then this implies that X is also complete. This is because X is made up of all possible closed balls within its space, so if each of these balls is complete, then there are no missing or "escaping" points in X as a whole.

Is the converse statement true - if X is complete, are all closed balls in X also complete?

No, the converse statement is not necessarily true. Just because X is complete does not mean that every closed ball within it is complete. Counterexamples can be found in spaces such as the rational numbers, where the entire space is complete, but not all closed balls within it are complete.

What is the significance of the completeness of closed balls in a space?

The completeness of closed balls in a space is an important property for many mathematical and scientific applications. It ensures that all possible Cauchy sequences within the space will converge to a point within the space, making it easier to analyze and make conclusions about the behavior of the space.

Are there any other equivalent statements to the statement "Every closed ball in X is complete, then X is complete"?

Yes, this statement is equivalent to the statement "Every closed subset of X is complete, then X is complete". This means that if all subsets of X are complete, then X as a whole is complete. It is also equivalent to the statement "Every Cauchy sequence in X converges to a point in X", as a Cauchy sequence in X would also be contained within a closed ball in X.

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