- #1
FallenApple
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Say I have a disk in ##R^2##. How would I know if it is compact? I mean, if the disk has no boundary, then we can have a limit that is outside the set. On the other hand, a disk with a boundary contains all limit points. But this seems unsatisfactory as for the open disk, we are assuming that the limit is outside the set. If we ignore the embedding space and look only at the universe of the disk, then one wouldn't even be able to speak of approaching a point outside the set.
So I tried to analyze it from using the definition of compact set.
If the set of all points in the disk is in the union of some finite number of open sets, which holds for every open cover of the disk, then by definition, the set is compact.
But I don't see how this would account for whether the disk has a boundary or not. We know a open disk isn't compact intuitively. Yet, how does the boundary arise from the definition of compactness? I mean, all we have are all the possible finite subcovers of the disk. We have nothing but open sets covering the set. But how would a closed set such as the closed disk arrive from such a definition. What is the point of having finite covers then?
So I tried to analyze it from using the definition of compact set.
A subsetof a topological spaceis compact if for every open cover ofthere exists a finite subcover of.
If the set of all points in the disk is in the union of some finite number of open sets, which holds for every open cover of the disk, then by definition, the set is compact.
But I don't see how this would account for whether the disk has a boundary or not. We know a open disk isn't compact intuitively. Yet, how does the boundary arise from the definition of compactness? I mean, all we have are all the possible finite subcovers of the disk. We have nothing but open sets covering the set. But how would a closed set such as the closed disk arrive from such a definition. What is the point of having finite covers then?
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