# [SOLVED]Compact Set Question

#### Sudharaka

##### Well-known member
MHB Math Helper
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

##### Well-known member
MHB Math Helper
I agree with you. However, the set $$\{x_j\}_{j= 0}^\infty$$ is compact. Could that "j= 1" be a printing error?

#### Sudharaka

##### Well-known member
MHB Math Helper
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.

#### Plato

##### Well-known member
MHB Math Helper
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

##### Well-known member
MHB Math Helper
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....

Thanks for pointing that out.