Since x n is bounded m r such that x n m for all n n

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Since ( x n ) is bounded, M R such that | x n | ≤ M for all n N . By definition, we have that s n x n ≥ - M . Therefore, since n N was arbitrary, ( s n ) n N is bounded below by - M and decreasing and we conclude that S := inf s n = lim n →∞ s n . We will inductively construct a subsequence ( x n k ) k N which converges to S . The first element Simply take x n 1 = x 1 . The k -th element ( k > 1 ) Assume we constructed x n k - 1 . We want to find x n k such that | x n k - S | < 1 /k . Since, S = lim m →∞ s m , M 0 k N such that | s m - S | < 1 / (2 k ) for any m M 0 k . In particular, we can find M k > n k - 1 such that | s M k - S | < 1 / (2 k ). Now since s M k = sup n M k x n , n k M k > n k - 1 such that s M k x n k < s M k + 1 / (2 k ). Then, | x n k - S | ≤ | x n k - s M k | + | s M k - S | < 1 / (2 k ) + 1 / (2 k ) = 1 /k. Since the sequence ( x n k ) k N satisfies | x n k - S | < 1 /k for all k , by the Archimedean property, > 0 , K ( ) N such that 1 /K ( ) < ⇒ | x n k - S | < k K ( ) and we conclude x n k S as k → ∞ .
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11. Let L R . The set L is said to be open if for any x L there exists > 0 such that ( x - , x + ) L . The set L is said to be closed if its complement L c = { x R : x / L } is open. (a) Prove that L is closed if and only if for any convergent sequence ( x n ) with x n L , the limit x = lim n →∞ x n = x is also an element of L . (b) Let ( x n ) be a bounded sequence. A point x R is called an accumulation point of ( x n ) if there exists a subsequence ( x n k ) of ( x n ) such that lim k →∞ x n k = x . We denote by L the set of all accumulation points of ( x n ). By the Bolzano-Weierstraß Theorem, the set L is non-empty. Prove that L is a bounded closed set. (c) Let ( x n ) be a bounded sequence, let L be as in part (b) and let S be as in problem 1. Prove that S = sup L . Solution: (a)( ) Let L be a closed set. Let ( x n ) n N be converging sequence with x n L and lim n →∞ x n = x . Then, in particular, > 0, n N such that | x n - x | < . Therefore, > 0, can find x n L such that x n ( x - , x + ). Therefore, > 0, ( x - , x + ) 6⊆ L c . Equivalently, @ > 0 such that ( x - , x + ) L c . Since L c is open, we must conclude that x / L c and therefore that x L . ( ) Let L be a set that is not closed. Then L c is not open, which implies that x L c such that n N , ( x - n - 1 , x + n - 1 ) 6⊆ L c ( x - n - 1 , x + n - 1 ) L 6 = . Hence, n N , x n L such that x n ( x - n - 1 , x + n - 1 ). We have constructed a sequence ( x n ) n N in L with the property that n N , | x n - x | < n - 1 , which implies that lim n →∞ x n = x where x / L . Therefore, it is not true that for any converging sequence ( x n ) n N with x n L , the limit x = lim n →∞ x n is also an element of L . Remark: A lot of people did well on the first implication. For the second, a lot of people have confused L is not closed” and “ L is open”: they do not mean the same thing. Some sets are both open and closed (clopen) and some sets are neither closed nor open.

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