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Let O be an open cover for [a,b] and let x=sup{d|[a,d] can be covered by finitely many elements of O}. Clearly x>a.

If x<b or x=b, then there is an element [tex]O_1[/tex] of O which is a neighborhood of x, and there is an [tex]\epsilon>0[/tex] such that [tex]x-\epsilon[/tex] is an element of [tex]O_1[/tex], and since there is a finite open subcover of O for [a,[tex]x-\epsilon[/tex]] (since x is the supremum), then adding [tex]O_1[/tex] to this subcover would form a finite subcover for [a,y] form some y>x, contradicting the idea that x is the supremum. Therefore, x>b.

2.

Consider a sequence in [a,b], and divide it in half, and choose the one with an infinite number of elements of the sequence. Divide it it in half again, and choose the one with an infinite number of elements of the sequence, and repeat this process to get a sequence of intervals. Due to completeness, their intersection is non-empty, which is therefore a limit point of the sequence. Therefore, [a,b] is sequentially compact, and therefore compact (since the space is second countable and therefore has the Lindelöf property).

The second proof is the process that I see used to prove the more general Heine Borel theorem for metric spaces - with an analogous process of picking smaller and smaller neighborhoods which are in turn totally bounded, and using the completeness property to deduce that there exists a limit point, and is therefore sequentially compact, and therefore is compact (since, due to total boundedness, there exists a set whose closure is the whole space, and therefore is second countable and therefore has the Lindelöf property). However, I am unable to find how to relate the first proof to the second.

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# How are these two proofs for the compactness of [a,b] equivalent?

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