Recall that N is the set of positive integers and Q is the set of rational numbers. NOTE: "Limit point\Sequences - Part 2 (Week 4) Infinite Limits Definition. A sequence (sn) is said to diverge to + and we write lim (sn) = + provided that for every M in R there exists a number N such that n > N implies that sn > M. Similarly, (sn) is said to diverge to - and we write lim (sn) = -, provided that for every M in R there exists a number N such that n > N implies that sn < M. Theorem. Suppose (sn) and (tn) are sequences such that sn tn for all n in N. (a) If lim (sn) = +, then lim (tn) = +. (b) If lim (tn) = -, then lim (sn) = -. *Theorem. Let (sn) be a sequence of positive numbers. Then lim (sn) = + iff lim (1/sn) = 0. Theorem. (a) If (sn) is an unbounded increasing sequence, then lim (sn) = +. (b) If (sn) is an unbounded decreasing sequence, then lim (sn) = -. Proof of (a): Suppose (sn) is an unbounded increasing sequence. Then S = {sn : n in N} is an unbounded set. Since the sequence is increasing, s1 is a lower bound for S, and since S is unbounded, S must be unbounded above. Given any real number M, there must exist an element of S larger than M. That is, there must exist integer N such that sN > M. But since (sn) is increasing, for all n > N, we have sn > sN > M. Therefore (sn) diverges to . The proof of (b) is similar. MORE ON SUBSEQUENCES: Recall: Definition. Let (sn) be a sequence. Let (nk) be a sequence of natural numbers such that n1 < n2 < n3 < .... The sequence \u0001\u0002\u0003\u0004 \u0005\tis called a subsequence of (sn). Theorem. [Prop. 2.1.17, p. 49 Lebl] If a sequence (sn) converges to a real number s, then every subsequence of (sn) also converges to s. Corollary: If two subsequences of (sn) converge to different numbers, then the sequence (sn) diverges. Page 1 of 7 Bounded Sequences Definition. Let (sn) be a bounded sequence. A subsequential limit of (sn) is any real number that is the limit of some subsequence of (sn) . There are several equivalent ways to define the limit superior and the limit inferior of a sequence. Here is one way (a way that is simpler to understand than Lebl, but equivalent): If S is the set of all subsequential limits of a bounded sequence (sn) , then we define the limit superior (or upper limit) of (sn) to be lim sup sn = sup S. Similarly, the limit inferior (or lower limit) of (sn) is lim inf sn = inf S. Example: \u0002\u0003 = \b1\u000b\u0003 + \u0003. \u0011 \u0012 \u0013 \u0014 \u0015 \u0016 \u000e0, \u0012 , \u0011 , \u0014 , \u0013 , \u0016 , \u0015 , ... \u0018 |\u0002\u0003 | = \u001a\b1\u000b\u0003 + \u001a |\b1\u000b\u0003 | + \u001a \u001a = 1 + 2 so the sequence is bounded. \u0003 \u0003 \u0003 For n odd, \u0002\u0003 = 1 + \u0003 1 and for n even, \u0002\u0003 = 1 + \u0003 1. There are two subsequential limits, -1 and 1, so S = {-1, 1}, lim inf sn = -1 and lim sup sn = 1. Here is an equivalent characterization of limit superior, which is essentially the definition of limit superior used in Lebl: Theorem. Let (sn) be a bounded sequence and let m = lim sup sn. Then the following properties hold: (a) for every > 0 there exists a number N such that n > N implies that sn < m + . (b) For every > 0 and for every positive integer i there exists an integer k > i such that sk > m - . Furthermore, if m is a real number satisfying (a) and (b), then m = lim sup sn. Corollary: Let (sn) be a bounded sequence and let m = lim sup sn. Then m S, where S is the set of subsequential limits of (sn). That is, there exists a subsequence of (sn) that converges to m. Page 2 of 7 \u0012 \u0012 \u0011 \u0011 \u0014 \u0014 Example: \b\u0002\u0003 = \u000e , , , , , , , , , , , , ... \u0018 \u0012 \u0012 \u0012 \u0011 \u0011 \u0011 \u0014 \u0014 \u0014 \u0013 \u0013 \u0013 Subsequences: \u0012 \u0012 \u0011 \u0011 \u0014 \u0014 \u0013 \u000e\t , , , , ... \u0018 converges to -1. \u0012 \u0011 \u0014 \u0013 \u000e , , , , ... \u0018 converges to 0. \u0012 \u0011 \u0014 \u000e\t\u0012 , \u0011 , \u0014 , \u0013 , ... \u0018 converges to 1. Set of subsequential limits = S = {-1, 0, 1}. lim inf sn = -1 and lim sup sn = 1. Be careful! Do not confuse a sequence (which is a function having domain N) with its range, which is a subset of the real numbers. Sequence values are put in a specified order but elements of a set are not. \b\u001e \u001f Example: \b\u0002\u0003 = \u000e \u0003 \u0018 = \u000e1, , , , , , ... \u0018 \u0012 \u0011 \u0014 \u0013 \u0016 The sequence converges to 0, so all subsequences converge to 0, and lim inf sn = 0 = lim sup sn . The range of the sequence is the set = !1, \u0011 , \u0013 , ... " !\u0012 , \u0014 , , ... ". \u0016 Notice that inf T = 1 and sup T = 1/2. Do not confuse the infimum and supremum of the range with the limit inferior and the limit superior of the sequence. Theorem. Suppose that (rn) converges to a positive number r and (sn) is a bounded sequence. Then lim sup rn sn = r lim sup sn. Proof: Suppose that (rn) converges to a positive number r and (sn) is a bounded sequence. Let s = lim sup sn and t = lim sup rn sn. We want to show that t = rs. Some subsequence \u0001\u0002\u0003\u0004 \u0005 must converge to s. Since (rn) converges, the subsequence \u0001$\u0003\u0004 \u0005 must converge to r. So, lim \u0001$\u0003\u0004 \u0002\u0003\u0004 \u0005 = lim $\u0003\u0004 lim \u0002\u0003\u0004 = rs. Therefore a subsequence of (rn sn) converges to rs, so rs lim sup rn sn = t. Since t = lim sup rn sn, some subsequence \u0001$\u0003\u0004 \u0002\u0003\u0004 \u0005 must converge to t. Since (rn) converges to a positive number r, we must have $\u0003\u0004 > 0 for nk sufficiently large. lim \u0002\u0003\u0004 =lim &\u001f\u0004 '\u001f\u0004 &\u001f\u0004 ( =& ( so & \tlim sup sn = s and we have t rs. Since rs t and t rs, we conclude that t = rs. Page 3 of 7 *Theorem. (Bolzano-Weierstrass Theorem for sequences) Every bounded sequence has a convergent subsequence. Proof: Let (sn) be a bounded sequence. Then the range of the sequence, S = { sn: n in N} is a bounded set. Case 1: S is finite. In this case, some value s in the sequence must be repeated infinitely many times. Then (s, s, s, ...) is a subsequence of (sn) which converges to s. Case 2: S is infinite. Since S is infinite and bounded, by the *Bolzano-Weierstrass Theorem for Sets , S has an accumulation point (limit point). * (SEE notes on the topology of real numbers and compact sets) Call the accumulation point s. Claim: There is a subsequence of (sn) which converges to s. Proof of claim: Since s is an accumulation point, any neighborhood of s must contain infinitely many values of S, i.e., infinitely many sequence values. For each positive integer k, consider the neighborhood N(s, 1/k)= {x: |x - s| < 1/k}. Choose \u0002\u0003) in N(s, 1). Choose \u0002\u0003* in N(s, 1/2) such that n2 > n1 (possible since there are infinitely many sequence values in the neighborhood.) In general, choose snk in N(s, 1/k) such that nk > nk - 1. We have constructed a sequence \u0001\u0002\u0003\u0004 \u0005\tsuch that for | \u0002\u0003\u0004 - s| < 1/k for all positive integers k. Since lim 1/k = 0, we can conclude that \u0001\u0002\u0003\u0004 \u0005\tconverges to s. Unbounded Sequences *Theorem. Every unbounded sequence contains a monotone subsequence that has either + or as a limit. Proof: Suppose (sn) is unbounded above. Claim: There is an unbounded increasing subsequence of (sn), and so a subsequence that diverges to +. Since (sn) is unbounded above, given any real number M, there must be infinitely many terms of (sn) which are larger than M. Choose \u0002\u0003) such that \u0002\u0003) > 1. Choose \u0002\u0003* such that \u0002\u0003* > 2 and \u0002\u0003* > \u0002\u0003) . In general, choose \u0002\u0003\u0004 such that \u0002\u0003\u0004 > + and \u0002\u0003\u0004 > \u0002\u0003\u0004,) . We have an increasing sequence \u0001\u0002\u0003\u0004 \u0005\tand for any integer M, \u0002\u0003\u0004 > \u0002\u0003- > M for all k M, so \u0001\u0002\u0003\u0004 \u0005\tis unbounded. (The proof is similar for a sequence (sn) which is unbounded below.) Page 4 of 7 Unbounded sequences: If (sn) is unbounded above, then define lim sup sn = +. If (sn) is bounded above but no subsequence converges to a finite number, define lim sup sn = . \u00032 \u0012 Example: \u0002\u0003 = .\tsin\u0012 \u000e \u0018. \b\t1,0,3,0,5,0,7, ... unbounded above Subsequences: \b\t0,0,0, ... converges to 0 \b\t1,3,5,7, ... diverges to + lim inf sn = 0 and lim sup sn = +. Cauchy Sequences Definition. A sequence (sn) of real numbers is said to be a Cauchy sequence if for each > 0 there exists a number N such that m, n > N implies that | sn - sm | < . Lemma. Every convergent sequence is a Cauchy sequence. Discussion: Suppose sequence (sn) converges to s. Suppose is any positive real number. We want to find N such that if m, n > N, | sn - sm | < . | sn - sm | = | sn - s + s - sm | adding and subtracting sn | sn - s | + | s - sm | by the triangle inequality Now use the fact that (sn) converges to s. Formal Proof: Suppose sequence (sn) converges to s. Suppose is any positive real number. Applying the limit definition, there exists N such that k > N implies that | sk - s | < /2. Then m, n > N implies that | sn - sm | = | sn - s + s - sm | adding and subtracting sn | sn - s | + | s - sm | by the triangle inequality < /2 + /2 = Therefore, (sn) is a Cauchy sequence. Lemma. Every Cauchy sequence is bounded. Proof similar to the proof that any convergent sequence is bounded. Page 5 of 7 *Theorem. (Cauchy Convergence Criterion). A sequence of real numbers is convergent if and only if it is a Cauchy sequence. Proof: Suppose (sn) converges. By the first lemma above, (sn) is a Cauchy sequence. Suppose (sn) is a Cauchy sequence. Let S = {sn : n in N}. By the preceding lemma (sn) is bounded, so S is bounded. Either S is a finite set or an infinite set. We will consider the two cases separately. Case 1: Suppose S is a finite set. If S consists of just one number s, then all of the sequence values are the same, and so the sequence converges to s. If S contains more than one number, let = min{ | sn - sm | : sn and sm are distinct values }, and so > 0. Since is Cauchy, there exists N such that n, m > N implies | sn - sm | < . But the only way that | sn - sm | can be less than the minimum distance is if | sn - sm | = 0 for all n, m > N. But this means sn - sm = 0 for all n, m > N, so sn = sm = some value s, so the sequence (sn) converges to s. Case 2: Suppose S is an infinite set. Since S is also bounded, by the Bolzano-Weierstrass Theorem, S must have an accumulation point, s, which is a real number. Claim: (sn) converges to s. By a homework exercise (topology of the real numbers section), every neighborhood of accumulation point s contains infinitely many points of S. Let be any positive real number. Then the neighborhood N(s; /2) contains infinitely many points of S. Therefore, there must be some N1 such that m > N1 implies sm N(s; /2). Also, since (sn) is Cauchy, there must be some N2 such that n, m > N2 implies that| sn - sm | < /2 Therefore, for any n > N = max{N1, N2}, | sn - s| = | sn - sm + sm - s | adding and subtracting sm | sn - sm | + | sm - s | by the triangle inequality < /2 + /2 = Therefore (sn) converges to s. Page 6 of 7 \u0012 \u0011 \u0014 *Definition The harmonic series is the infinite series 1 + + + + = 8 \u00039 \u0003 Example. The harmonic series diverges to . \u0012 \u0011 \u0003 Let \u0002\u0003 = 1 + + + + . If m > n, then \u0002: \u0002\u0003 = > \u0003; + \u0003;\u0012 : : ++ + ++ : : There are m - n terms with m - n terms = \b< . : \u0003 =1: If m = 2n, then we have \u0002\u0012\u0003 \u0002\u0003 = 1 \u0012 = \u0012. So, no matter how far out in the sequence we go, we can find sequence values that are more than 1/2 apart. Therefore, (sn) is not a Cauchy sequence and therefore diverges. Clearly (sn) is an increasing sequence and now we know it diverges, so (sn) diverges to . That is, lim sn = 1 + \u0012 + \u0011 + \u0014 + = \u00039 \u0003 = . Page 7 of 7 Topology of the Real Numbers ; Compact Sets See Section 1.3 of Trench. Definition: Let x R and let > 0. A neighborhood of x (or an -neighborhood of x) is a set of the form N(x; ) = {y R : |x - y| < }. The number is referred to as the radius of N(x; ). x- x x+ N(x; ) is the open interval (x - , x + ), which is centered at x and has width 2 (so radius ). A deleted neighborhood of x is a set of the form N*(x; ) = {y R : 0 <|x - y| < }. Note that N*(x; ) = N(x; ) \\ {x}. N*(x; ) is the union of two open intervals, (x - , x) (x, x + ). x- x x+ Let S be a subset of R. A point x in R is an interior point of S if there exists a neighborhood N of x such that N S. If for every neighborhood N of x, N S and N (R \\ S) , then x is called a boundary point of S. That is, if every neighborhood of x contains at least one point in S and at least one point not in S, then x is a boundary point. The set of all interior points of S is denoted int S or So (Trench) and the set of all boundary points of S is denoted bd S or S (Trench). Examples: (a) For any open interval S = (a, b), every point in the interval is an interior point. Proof: Say that x is in (a, b). We compute the distances between x and the endpoints and take the smaller distance: Let = min { x - a, b - x}. Then N(x; ) (a, b), so x is an interior point. Therefore, int S = S. x-a a b-x x b Also, bd S = {a, b}. Any neighborhood of an endpoint contains points in the interval (a, b) and points outside (a, b). (b) For any closed interval T = [a, b], int T = (a, b) and bd T = {a, b}. (c) For a half-open interval of the form V = [a, b), int V = (a, b) and bd V = {a, b}. (d) For an infinite interval of the form W = (-, b], int W = (-, b) and bd V = {b}. (e) For the set R of all real numbers, int R = R and bd R = . Let S be a subset of R. There are several equivalent ways to define "open set" and "closed set". Characterizations of "open": A set S is open: iff S = int S. "The set's interior is the set itself". iff bd S R \\ S. "The set's boundary points are contained in the set's complement." Characterizations of "closed": A set S is closed iff the complement R \\ S is open. "The set's complement is open". iff bd S S "The set contains all its boundary points." Page 1 of 7 Examples: (a) Interval (a, b) is open. (b) Interval [a, b] is closed. (c) Interval [a, b) is neither open nor closed. (d) Interval (-, b] is closed. (e) The set Q of rational numbers is neither open nor closed, because every open interval contains both rational numbers and irrational numbers. Every real number is a boundary point of Q. (f) The set R of real numbers is both open and closed. The empty set is both open and closed. Theorem. (a) The union of any collection of open sets is an open set. [Trench, Th 1.3.3(a)] (b) The intersection of any finite collection of open sets is an open set. Proof: (a): Suppose C be a collection of open sets and let \u0001 = \u0005\u0007 \u0004 If x S, then x A for some A in C . Since A is open, x is an interior point of A. So, there exists a neighborhood N of x such that N A. But A S, so N S and x is an interior point of S. Hence S is open. (b) Let A1, A2, ..., An be a finite collection of open sets and let = \u000e \u0004 If T is empty, then we are done, since the empty set is open. If T is nonempty, suppose x T. Then x Aj for all j = 1, 2, ..., n. Since each set Aj is open, there exist neighborhoods Nj(x; j) of x such that Nj(x; j) Aj. Let = min {1, 2, ..., n}. Then N(x; ) Aj for each j = 1, ..., n, so N(x; ) T. Thus x is an interior point of T, and T is open. Corollary. (a) The intersection of any collection of closed sets is closed. [Trench, Th 1.3.3(b)] (b) The union of any finite collection of closed sets is closed. Proof: (a): Suppose C is a collection of closed sets and let \u0001 = \u0010\u0007 \u000f. To show that S is closed is equivalent to showing that its complement R \\ S is open. But \u0011\\\u0001 = \u0011\\ \u0010\u0007 \u000f = \u0010\u0007\t(\u0011\\\u000f) by De Morgan's law for sets. Since B is closed, R \\ B is open. By part (a) of the previous theorem, the union of a collection of open sets is open. Therefore R \\ S is open, which is what we wanted to show. The proof of part (b) uses a similar approach. Page 2 of 7 Example: In part (b) of the theorem, it matters that we have a finite collection of open sets. If we have an infinite collection, then the infimum of the epsilons in the proof could be 0, and we might not have a neighborhood of positive radius fitting inside all of the open sets. For instance, if Aj = (-1/j, 1/j) for j N, all of the intervals are open but the intersection \u000e \u0004 = \u00150\u0017 , a closed set. Def. Let S be a subset of R. A point x in R is an accumulation point of S [or called a limit point (Trench) or a cluster point (Lebl)] iff every deleted neighborhood of x contains a point of S. That is, for every > 0, N*(x; ) S . Or, equivalently, every neighborhood of x contains a point of S other than x. An accumulation point of S is not necessarily contained in S. The set of all accumulation points (i.e., limit points) of S is denoted S'. If x S and x S', then, x is called an isolated point of S. Examples: (a) For any open interval S = (a, b), the set of limit points is S' = [a, b]. (b) If S = {1/n | n N}, then the set of limit points S' = {0}. Note that the limit point is not in the set S. (c) For the set of natural numbers, N' = . All of the natural numbers are isolated points. We can separate each natural number by enclosing it in a neighborhood of radius 1/4. (d) For any finite set S, S' = . Characterizations of "closure of a set": The closure of S, denoted cl S or \u0019 \u0018: (Trench) cl S = S (bd S) The closure of set S is the union of S and its boundary points. (Trench) cl S = S S' The closure of set S is the union of S and its accumulation (limit) points. Theorem. Let S be a subset of R. Then (a) S is closed iff S contains all of its accumulation (limit) points. (b) cl S is a closed set (c) S is closed iff S = cl S Proof: (a): Suppose S is closed. Then R \\ S is open and int(R\\S) = R\\S. Suppose x S'. We want to show that x S, so we can conclude that S' S. Since x S', every neighborhood of x contains a point of S other than x. Therefore, no neighborhood of x is contained within R \\ S, the complement of S. So, x int (R \\ S) = R \\ S. Since x is not in the complement of S, x S. Suppose that S contains all of its accumulation points, that is, S' S. We want to show that S is closed, or equivalently, that R \\ S is open. Let x R \\ S. Since S' S, x S'. So, some neighborhood of x does not contain a point in S. Therefore this neighborhood must be contained in R \\ S, the complement of S. So x int(R \\ S), and R \\ S is open. Page 3 of 7 Compact Sets Definitions. If F is a collection of open sets whose union contains S, then F is called an open cover of S. If G F and G is also an open cover of S, then G is called a subcover of S. A set S is compact if every open cover of S contains a finite subcover. Examples: (a) Let S = (0, 2) . Not compact The collection of open intervals An = (1/n, 3) is an open cover of (0, 2) which has no finite subcover. Compact. (b) Let S = { x1, x2, x3, ... xn } be a finite subset of R. Suppose F is any collection of sets which form an open cover of S. Since S is contained in the open cover, for each xi in S, there must be a corresponding set (denoted Ai) in the cover F which contains xi. Then {A1, A2, A3, ..., An} is a finite subcover. (c) Let S = [0, ) . Not compact The collection of open intervals An = (-1, n) is an open cover of [0, ) which has no finite subcover. * Lemma: If S is a nonempty closed bounded subset of R, then S has a maximum and a minimum. Proof: Suppose S is a nonempty closed bounded subset of R. Since S is bounded above, by the Completeness Axiom, S has a supremum. Call it m. We want to show that m is in S. Since m is the least upper bound, for every > 0, there exists x S such that m - < x (because m is the least upper bound) and x < m (since m is an upper bound). So, m - < x < m. This means that each deleted neighborhood N*(m, ) contains a point of S, and therefore m is an accumulation point of S. Since S is closed, S contains all of its accumulation points, and so m must be in S. Since m = sup S and m S, m is the maximum of the set. The proof involving the minimum is very similar. Page 4 of 7 *Theorem (Heine-Borel): A subset S of R is compact iff S is closed and bounded. Proof: Suppose S is compact. We must show that S is closed and bounded. First we will show that S is bounded. For each n N, let In = N(0; n) = (-n , n). Then each In is open and\t\u0001\t \u001b \u000e \u001a , so {In, n N} is an open cover of S. Since S is compact, there is a finite subcover , { In1, In2, ..., Ink}. Let m = maximum of {n1, n2, ..., nk}. Then | x | < m for all x S, and so S is bounded. We must still show that S is closed. Suppose S is NOT closed. (We will derive a contradiction.) Then there must be an accumulation point p in the closure of S but not in S, i.e., p in (cl S) \\ S. Consider the collection of closed intervals centered at p with radius 1/n : Let Cn = [p - 1/n, p + 1/n]. Note that \u001b \u000e \u001c = \u0015\u001d\u0017 and C1 C2 C3.... Let Un = R \\ Cn. Then Un is open and U1 U2 U3 .... \u001b \u001b \u000e \u001e = \u000e(\u001f\\\u001c ) = \u001f\\ \u001b \u000e \u001c by De Morgan's law for sets = R \\ {p} S, since p is not in S. Therefore {Un, n in N} is an open cover of S and since S is compact, there must be a finite subcover , { Un1, Un2, ..., Unk}. Since the "U" sets are nested, the union of the sets in the finite subcover is just the largest set Unk. and so, S must be contained within this one set; i.e., S Unk. S Unk But Unk. =R \\ Cnk = R \\ [p - 1/nk, p + 1/nk]. p-1/nk p p+1/nk So S is contained in the complement of the interval [p - 1/nk, p + 1/nk]. Therefore S and [p - 1/nk, p + 1/nk] have no points in common. However, this means we have a neighborhood of p which does not contain any points of S. Then p cannot be an accumulation point, which contradicts our definition of p at the start. Therefore S must be closed. Page 5 of 7 Proof of the converse: Suppose that S is closed and bounded. We must show that S is compact. Let F be any open cover of S. If S is empty, then S is vacuously compact and we are done. Suppose S is nonempty. Applying the lemma, since S is closed and bounded, S must have a minimum m and a maximum M. (Set S depicted in red) m M Let Sx = S (-, x]. Sets of form (-, x] in blue. [ [ Note that the sets Sx are nested. When x = m, Sm = S (-, m] = {m} since m is the minimum of S. As x increases, Sx encompasses more and more elements of S, until we get to x = M. SM = S (-, M]= S, since M is the maximum. For any x value larger than M, Sx = S (-, x] is still S. Now let B = {x: Sx is covered by a finite subcover of F }. The minimum of S, m, must be in B, because Sm = {m}, and m is covered by any single set in the cover F which contains m. Therefore B is nonempty. Claim: B is not bounded above. (Shown later) If we can establish this claim, this means that M, the maximum of S must be contained in B. But by definition of B, saying that M is in B means that SM (= S) is covered by a finite subcover of F, and therefore S is compact. Proof of claim: Suppose not. Suppose that B is bounded above. Then B must have a least upper bound, sup B, which we will denote by b. Then either b is in S or not. If b is in S, since F is an open cover of S, there is a set F0 in the open cover such that b is in S. Since F0 is open there is a neighborhood (x1, x2) of b contained in F0. But since x1 < b, and b = sup B, we must have a finite open cover for Sx1. But then F0 together with the finite open cover forms a new finite subcover that covers Sx2 as well as Sb. But that means b is not the least upper bound, which is a contradiction. Therefore b cannot be in S. Page 6 of 7 If b is not in S, then b is in R \\ S. Since S is closed, R \\ S is open and there must be some neighborhood N(b; ) of b contained in R \\ S. That is, N(b,) S = . But N(b, ) = (b - , b + ) by definition of neighborhood. Sb- = S (-, b - ]. Sb+/2 = S (-, b +/2]. Since b - B, we must also have b + /2 B. But then b cannot be the supremum of B as supposed. So, if B has an upper bound b, it cannot be in S and it cannot be in R\\S either. Therefore, upper bound b cannot exist. B must be unbounded as we claimed. *Theorem (Bolzano-Weierstrass, for sets) If a bounded subset S of R contains infinitely many points, then there exists at least one point in R that is an accumulation point of S. Proof in video Theorem. Let F = {K : A} be a family of compact subsets of R. Suppose that the intersection of any finite subfamily of F is nonempty. Then {K : A} \t. Proof in video *Corollary (Nested Intervals Theorem). Let F = {An : n N} be a family of closed bounded nested intervals in R such that A1 A2 A3.... Then \u001b \u000e \u0004 . Proof in video Page 7 of 7