Totally bounded & Heine Borel?

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SUMMARY

The discussion clarifies that a set being totally bounded does not imply it possesses the Heine-Borel property. Specifically, the Heine-Borel property requires that every open cover has a finite subcover, while totally bounded sets allow for the selection of covers based on a fixed radius. An example provided is the set (0,1), which can be covered by an infinite collection of open balls B(x,1/n) for n>1, demonstrating that no finite subcover exists. This distinction is crucial for understanding compactness in metric spaces.

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Homework Statement


Why doesn't a set being totally bounded imply the set has the Heine Borel property?

Another related question is what happens if a cover consist of open balls that cover the set and more of it? i.e. A=(-1,2) U (1,3) covers (0,1) but really covers more than (0,1). Is A considered a legitimate cover? I have a feeling it is.

The Attempt at a Solution


They both refer to the set being covered by a finite number of sets.

Is it because for example, Take the set (0,1). One cover for it is an infinite cover consisting of infinitely many open balls of the type B(x,1/n), n>1. There is no finite subcover of this cover that covers (0,1). But such a cover is not allowed to show totally boundedness as the radius has to be fixed.
 
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What is the Heine Borel property precisely?

The difference is that for compactness, which I assume is H-B, the cover is given to you: you have no choice over the size of the sets, or where they are centred. For totally bounded, you get to choose how to lay down the cover once the radius has been fixed.
 

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