By Min Yan

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**Additional info for Advanced Analysis**

**Example text**

Suppose f (x) is an increasing function defined and unbounded for x > a and x near a. Prove that limx→a+ f (x) = −∞. Finally, the Cauchy criterion can also be applied to the limit of functions. 10. A function f (x) has limit at a if and only if for any > 0, there is δ > 0, such that 0 < |x − a| < δ, 0 < |y − a| < δ =⇒ |f (x) − f (y)| < . 3. LIMIT OF FUNCTION 59 Similar statements can be made for the one side limits and the limits at infinities. Proof. 2 can be adopted without much change. Conversely, assume f (x) satisfies the Cauchy condition.

We denote the corresponding interval by I1 = [α1 , β1 ] and denote X1 = X ∩ I1 . α1 + β1 Further divide I1 into two equal halves I1 = α1 , and I1 = 2 α1 + β1 , β1 . Then either X1 = X1 ∩ I1 or X1 = X1 ∩ I1 cannot be covered 2 by finitely many open intervals in U. We denote the corresponding interval by I2 = [α2 , β2 ] and denote X2 = X ∩ I2 . Keep going, we get a sequence of intervals I = [α, β] ⊃ I1 = [α1 , β1 ] ⊃ I2 = [α2 , β2 ] ⊃ · · · ⊃ Ik = [αk , βk ] ⊃ · · · β−α with the length of Ik being βk − αk = , and Xk = X ∩ Ik cannot be 2k covered by finitely many open intervals in U.

Suppose X ⊂ I = [α, β] for a bounded and closed interval I. Suppose X cannot be covered by finitely many open intervals in U = {(ai , bi )}. Similar to the proof of Bolzano-Weierstrass Theorem, we divide the inα+β α+β terval into two equal halves I = α, and I = , β . Then 2 2 either X = X ∩ I or X = X ∩ I cannot be covered by finitely many open intervals in U. We denote the corresponding interval by I1 = [α1 , β1 ] and denote X1 = X ∩ I1 . α1 + β1 Further divide I1 into two equal halves I1 = α1 , and I1 = 2 α1 + β1 , β1 .

### Advanced Analysis by Min Yan

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