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So all closed interval [a,b] is compact

(see Theorem 2.2.1 in Real Analysis and Probability by RM Dudley)

Now, Let's say I have [0,10] as my closed interval.

Let My Open Cover be

(0, 5)

(5, 7.5)

(7.5, 8.75)

(8.75, 9.375)

....

Essentially, The length of each open interval is cut by half, i.e.

length (0,5) = 5

length (5,7.5) = 2.5 = 5/2

length (7.5,8.75) = 1.25 = 5/(2*2)

length (8.75, 9.375) = 0.625 = 5/(2*2*2)

....

the Union of all these interval gives us [0.10] \ ({0} U {5} U {7.5} U {8.75} U {9.375} ... )

Therefore, we add

(0-[tex]\epsilon[/tex], 0+[tex]\epsilon[/tex])

(5-[tex]\epsilon[/tex], 5+[tex]\epsilon[/tex])

(7.5-[tex]\epsilon[/tex], 7.5+[tex]\epsilon[/tex])

(8.75-[tex]\epsilon[/tex], 8.75+[tex]\epsilon[/tex])

.... to cover all these missing single points.

The choice of [tex]\epsilon[/tex] is arbitrary, as long as it doesn't touch the middle-point of the next interval.

for example,

to cover {5}, we can add a (5-0.1, 5+0.1)

to cover {7.5} we can add a (7.5-0.01, 7.5+0.01)

etc etc

So we have create ourselves an Open cover for [0, 10]

but i can't see any finite members of such cover become yet another cover for [0, 10]

I know [0, 10] is compact, I simply can't find a finite subcover.

Where is the loophole ?

Thanks

J

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# Any closed interval [a,b] is compact ?

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