MHB Compact Set Question: Counterexample Proved

[Sudharaka]

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.

 

[HallsofIvy]

I agree with you. However, the set \{x_j\}_{j= 0}^\infty is compact. Could that "j= 1" be a printing error?

 

[Sudharaka]

I agree with you. However, the set \{x_j\}_{j= 0}^\infty 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.

 

[Plato2]

I agree with you. However, the set \{x_j\}_{j= 0}^\infty 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 {\lim _{n \to \infty }}d({a_n},{a_0}) \ne 0

 

[Sudharaka]

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 {\lim _{n \to \infty }}d({a_n},{a_0}) \ne 0

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

Thanks for pointing that out. :)