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

[deckoff9]

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.

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 n→∞. Is there something wrong with my logic or did the book screw up by not mentioning complete in the conditional?

 

[micromass]

Your analysis book is wrong (or you misinterpreted it). Which book are you using? Can you quote the statement.

The only thing we can say that any closed and bounded subset of $\mathbb{R}^n$ is compact. It does NOT hold for arbitrary metric spaces!
Indeed, ]0,1[ is closed and bounded in itself but not compact!

In arbitrary metric spaces, we got the statement: any complete and totally bounded set is compact.

 

[deckoff9]

Thanks for the quick answer micromass. Yeah that's what I thought. The book I'm using is Introduction to Analysis by Rosenlicht. The statement is exactly what I wrote, "Any closed bounded subset of E^n is compact." and the notation has been E is an arbitrary metric space.

 

[dextercioby]

Then I guess you need to change the book. Rudin I believe wrote a good book on real analysis.

 

[spamiam]

..."Any closed bounded subset of E^n is compact." and the notation has been E is an arbitrary metric space.

Are you sure about this notation? In my topology text, $\mathbb{E}$ denotes the real numbers with the topology induced by the Euclidean norm.

 

[lavinia]

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.

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 n→∞. Is there something wrong with my logic or did the book screw up by not mentioning complete in the conditional?

(),1) is closed in itself but no closed in R. Any topological space is closed in itself.

 

[deckoff9]

Since he used the fact E^n was complete in the proof I guess I'll give the benefit of the doubt and say maybe I misunderstood the notation. Anyways thanks for the helps~

 

[Sina]

Use Marsden Elementary Classical Analysis.

But it is strange, I don't think any author would do that big of a mistake.