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Unlike in $R^n$, closed and bounded doesn't guarantee sequential compactness. Textbook examples includes sup metric and R^infinite metric. I am wondering what would be a example of closed and bounded doesn't imply sequential compactness in more general Metric space $(X,\rho)$?

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Take the space $l_2$ (ie, $d(x,y) = \sqrt{\sum_n |[x]_n - [y]_n|^2}$) and the set $C = \{ e_n \}_n$, where $e_n$ is the $n$th unit vector (ie, zeros everywhere except for the $n$th position which is one).

Then $C$ is bounded, closed but not compact (or sequentially compact).

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Take any metric space $(X,\rho)$ which is not sequentially compact (= compact, for metric spaces), and replace $\rho$ with the standard bounded metric $\rho_1(x,y) = \min(\rho(x,y),1))$. Then $X$ itself is closed and $\rho_1$-bounded but not compact, despite the fact that the induced topology $\tau_{\rho_1}$ is the same as the induced topology $\tau_{\rho}$ of $(X,\rho)$.

I hope that this example drives home that the Heine-Borel Theorem is fundamentally a metric result and not a topological one.

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The closed unit ball in any infinite-dimensional normed space is not compact. This is an easy consequence of Riesz's lemma.

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