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- #1

- Feb 5, 2012

- 1,621

I encountered the following question recently.

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,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.

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

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.