Open cover with no finite subcover

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SUMMARY

The discussion focuses on finding an open cover for the interval (0,1) that lacks a finite subcover. The proposed open cover consists of intervals O_x defined as (x/2, 1) for each x in (0,1). It is established that any finite collection of these intervals cannot cover all points in (0,1) due to the existence of points less than n/2 not included in the finite union. In contrast, the closed interval [0,1] requires endpoints to be included in any open cover, allowing for a finite subcover when specific intervals are chosen.

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Homework Statement


I want to find an open cover of x>0 with no finite subcover.



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The Attempt at a Solution


What about (0,1). Would{1/n,1} have no finite subcover?
 
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Let x be a point in (0,1) and let O_x be the open interval (x/2, 1). Then the union from 0 to infinite of your O_x's covers (0,1). Now to find a finite subcover, consider a finite subcollection of O_x's. Try to see here how reducing the upper limit on your union from infinite to some finite value reduces the points that are contained in the cover. For example, let the finite subcover be the union of O_1, O_2,..., O_n. Then if y is the min of 1, 2,..., n, any number less than n/2 is not contained in the finite union.

To think about how this is different for closed sets, consider [0,1], the same thing as the previous set except that it contains 0 and 1. An open cover for this set must contain it's endpoints. So we can take an open cover for this set by fixing a c > 0 and let O_1 = (-c,c), O_2, = (1-c, 1+c), etc. Then you can see that this covers [0,1]. But it contains a finite subcover since now we can choose y such that y/2 is less than c. So the addition of the set (-c, c) allows us to take the finite subcover as the union of O_1, O_2, and O_y. Note that this is only one open cover of [0,1], but you can try other cases for examples.
 
Oh ok, I see how that works.
 

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