Compact Set Question: Counterexample Proved

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In summary, the set \(\{x_{j}\}_{j=1}^{\infty}\) is not compact as the limiting value of the sequence (and hence all subsequences) does not belong to \(\{\frac{1}{n}\}_{n=1}^{\infty}\).
  • #1
Sudharaka
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Hi everyone, :)

I encountered the following question recently. :)

Let \((X,\,d)\) be a metric space and \(\lim_{n\rightarrow\infty}d(x_n,\, x_0)=0\). Prove that the set \(\{x_{j}\}_{j=1}^{\infty}\) is compact.

Now I think this question is wrong. Let me give a counterexample. Take the set of real numbers with the usual Euclidean metric. Then take for example the sequence, \(\{\frac{1}{n}\}_{n=1}^{\infty}\). Then,

\[\lim_{n\rightarrow\infty}d(x_n,\, x_0)=\lim_{n\rightarrow\infty}\left|\frac{1}{n}-0\right|=0\]

All subsequences of \(\{\frac{1}{n}\}_{n=1}^{\infty}\) should converge to the same limit, which in this case is zero. Hence \(\{\frac{1}{n}\}_{n=1}^{\infty}\) is not compact as the limiting value of the sequence (and hence all subsequences) does not belong to \(\{\frac{1}{n}\}_{n=1}^{\infty}\). Let me know if I am wrong. :)

Thank you.
 
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  • #2
I agree with you. However, the set [tex]\{x_j\}_{j= 0}^\infty[/tex] is compact. Could that "j= 1" be a printing error?
 
  • #3
HallsofIvy said:
I agree with you. However, the set [tex]\{x_j\}_{j= 0}^\infty[/tex] is compact. Could that "j= 1" be a printing error?

Hi HallsofIvy, :)

Thanks for replying. But how can that make a difference? For example I can define a sequence in the real numbers like,

\[x_{0}=1\mbox{ and }x_n=\frac{1}{n}\mbox{ for }n\geq 1\]

which is again convergent to \(0\) but not compact.
 
  • #4
Sudharaka said:
HallsofIvy said:
I agree with you. However, the set [tex]\{x_j\}_{j= 0}^\infty[/tex] is compact. Could that "j= 1" be a printing error?
But how can that make a difference? For example I can define a sequence in the real numbers like,
\[x_{0}=1\mbox{ and }x_n=\frac{1}{n}\mbox{ for }n\geq 1\]
which is again convergent to \(0\) but not compact.

But now [tex]{\lim _{n \to \infty }}d({a_n},{a_0}) \ne 0[/tex]
 
  • #5
Plato said:
Sudharaka said:
But how can that make a difference? For example I can define a sequence in the real numbers like,
\[x_{0}=1\mbox{ and }x_n=\frac{1}{n}\mbox{ for }n\geq 1\]
which is again convergent to \(0\) but not compact.

But now [tex]{\lim _{n \to \infty }}d({a_n},{a_0}) \ne 0[/tex]

Arghaaaaaa... How could I missed that... (Angry) (Headbang)

Thanks for pointing that out. :)
 

FAQ: Compact Set Question: Counterexample Proved

What is a compact set?

A compact set is a subset of a metric space that is closed and bounded, meaning that it contains all of its limit points and has a finite diameter.

What is a counterexample?

A counterexample is an example that disproves a conjecture or statement by providing a specific case in which the statement does not hold true.

How is a counterexample used to prove a compact set question?

A counterexample can be used to prove that a set is not compact by providing a specific case in which the set does not satisfy the definition of a compact set. This disproves the statement that the set is compact.

Can a compact set question be proved without a counterexample?

Yes, a compact set question can also be proved by using the definition of a compact set and showing that the set satisfies all of the necessary criteria.

Why is it important to prove a compact set question?

Proving a compact set question is important because it allows us to understand the properties and behavior of compact sets, which have many applications in mathematics and other scientific fields. It also helps to build a stronger foundation for further research and exploration in the field.

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