Hello. In my analysis book, it says that "Any closed bounded subset of E^n is compact" where E is an arbitrary metric space. I looked over the proof and it used that fact that E^n was complete, but it does not say that in the original condition so I was wondering if the book made a mistake in not adding that.(adsbygoogle = window.adsbygoogle || []).push({});

For my counter example, consider the metric space (0,1), with the usual distance metric. The subset of itself is closed by definition, and it is bounded. However, it is not compact, since (1/n, 1-1/n) covers it as nāā. Is there something wrong with my logic or did the book screw up by not mentioning complete in the conditional?

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# Compactness of (0,1) when that is the whole metric space

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