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By Terence Tao

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Every constant sequence (xn bounded. The sequence Xn = (_1)n is bounded. 33 S. K. Berberian, A First Course in Real Analysis © Springer Science+Business Media New York 1994 x for all n) is 3. 4. Example. The sequence xn = n is unbounded. 5. Theorem. If (xn) and (Yn) are bounded sequences in IR, then the sequences (xn + Yn) and (xnYn) are also bounded. K Proof If IXnl : : : K and IYnl::::: K' then IXn + K' and IXnYnl = IXnllYnl : : : KK' . 0 + Ynl ::::: IXnl + IYnl : : : Exercises 1. Show that if (xn) and (Yn) are bounded sequences and c E IR, then the sequences (cxn) and (x n - Yn) are also bounded.

8). 3) . Note that A contains numbers > 0; for , if c ~ 1 then 1/2 E A (because 1/4 < 1 S c), whereas if 0 < c < 1 then c E A (because c2 < c). It follows that , > O. Next, we assert that , E B; by the arguments in the preceding section, we need only show that A has no largest element. Assuming a E A, let's find a larger element of A. If a S 0 then any positive element of A will do. Suppose a > O. We know that a 2 < C; it will suffice to find a positive integer n such that (a + 1/n)2 < c. 4 (with 2 replaced by c).

If m > 1 then m - 1 is a positive integer smaller than m , so it can't belong to S; this means that m - 1 ::; x , thus x E [m - 1, m) and n = m - 1 fills the bill. 2. Definition. With notations as in the theorem, the integer n is denoted [x] and the function IR --+ Z defined by x 1-+ [x] is called the bracket function (or the greatest integer function , since [x] is the largest integer that is ::; x ). Exercises 1. 3] and [-2]. 2. Sketch the graph of each of the following functions f: IR (i) f(x) = [x].

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