HW 4 SOLUTIONS. , x + x x 1 ) 2

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1 HW 4 SOLUTIONS The Way of Analysis p. 98: 1.) Suppose that A is open. Show that A minus a finite set is still open. This follows by induction as long as A minus one point x is still open. To see that A \ {x} is open, notice that (, x) (x, ) is the union of two open intervals and thus open; since the finite intersection of open sets is open, so is A \ {x} = A ((, x) (x, )). This is not true if we take away a countable set. For example, the set of irrational numbers R \ Q is not open since Q is dense (so every open interval about an irrational still hits Q). 2.) Let x n be a sequence and A the set whose elements are the x n s. Show that a L.P. of A is a L.P. of the sequence: If x is a L.P. of A, we get infinitely many points in A (x 1/n, x + 1/n) for every n. This means there are infinitely many terms of the sequence in A (x 1/n, x + 1/n) for every n. Hence, x is a L.P. of the sequence too. Show that if no point in A occurs more than a finite number of times in the sequence, then a L.P. of the sequence is also an L.P. of A: If z is a L.P. of the sequence, then every nbhd of z contains infinitely many x n s. Since each term in the sequence occurs only a finite number of times, infinitely many of these points are different so z is also an L.P. of the set. 4.) Let A be a set and x R. Show that x is a L.P. of A iff there is a sequence x n of distinct points in A that converges to x. Suppose first that there is a sequence x n of distinct points in A that converges to x. This means that every nbhd of x contains some tail of the sequence {x n }, and since the sequence consists of distinct points, this means there are infinitely many of the x n s in any nbhd.; hence x is a L.P. of A. Suppose now that x is a L.P. of A. By definition, the nbhd (x 1, x + 1) contains infinitely many points of A and, in particular, a point x 1 x belonging to A. Likewise, the nbhd (x x x 1, x + x x 1 ) 2 2 1

2 2 HW 4 SOLUTIONS contains a point x 2 x belonging to A. Notice that x 2 x 1! Now continue this procedure to define a sequence x n. It is not difficult to show that {x n } constructed in this way converges to x and you should do this. 5.) Let A be a closed set, x A, and B = A \ {x}. When is B closed? B is closed iff x is NOT a LP of A. To see this, simply note that x is a LP of A iff x is a LP of B (since removing any finite number of points from a set does not change its LP s this is obvious from the definition of a limit point p: there have to be INFINITELY MANY points of the set in any nbhd of p). 6.) Prove that every infinite set A has a countable dense subset: Consider the set of rational open intervals, that is all intervals of the form (a, b) where a, b Q. Notice (and argue!) that the set of all such intervals is a countable set (it s basically Q Q), list these: I 1, I 2,... For each I k, choose a point a k I k A if possible. This gives an (at most) countable set of points in A; we will show that it is dense. Suppose now that a A and I is some neighborhood of a. By the density of rationals, there is a rational interval I k I. Since I k A is nonempty (it at least contains a), we were able to choose a point a k I k A I A. Give an example of a set A such that the intersection of A with the rational numbers is not dense in A: This is easy - just let A be the irrationals! The intersection of this set with the rationals is empty - so it certainly is not dense in the irrationals. 9.) Given a closed set A, construct a sequence whose set of LP s is A: Let a k be the elements in the countable dense subset given by exercise 6 and consider the sequence a 1, a 1, a 2, a 1, a 2, a 3,... Since every a k appears infinitely often, each a k is an LP of the sequence; since the set of LP s is closed (by exercise 8) and the a k s are dense in A, every point of A is an LP of the sequence. We have not used yet that A is closed, but we do so now: we have shown that all points in A are limit points of the sequence above, but we have to still show that there are no limit points of the sequence above outside of A. Since A is closed, any Cauchy sequence of points in A converges to a point in A. Since the a k s are in A, every Cauchy subsequence of the above sequence must therefore converge to a point in A. Hence any limit point of the sequence must be in A. 10.) Show that the set of numbers of the form k/5 n (where k Z and n N) is dense in R. Suppose that x R. We will construct a sequence k n /5 n of these numbers converging to x. For each n, define k n to be the largest integer with k n 5 n x.

3 It follows that x is between kn 5 n and kn+1 5 n and therefore HW 4 SOLUTIONS 3 k n 5 x < 1 n 5. n This sequence { kn } converges to x since 1 0 as n. 5 n 5 n 14.) What sets are both open and closed? Just R and the empty set! These two sets are obviously both open and closed, so the problem is to show that they are the only sets which are both open and closed. Suppose therefore that A R is nonempty, open, and closed. Fix a point a A (since it s nonempty). Suppose that there is a point b R \ A; we will show that this leads to a contradiction. We can assume that b > a (the other case is similar). Define the set C = {x A a < x < b} = A (a, b). This is an intersection of two open sets and hence also open (and nonempty since a A!). Therefore, C does not contain the point sup C (if it did, it would contain an interval around it). However, since C A and A is closed, we have sup C A. Hence we must have that sup C is not in (a, b), and tt follows that sup C must equal b. This is a contradiction: since b / A, we have sup C < b. p. 106: 1.) Compact sets are closed under arbitrary intersections and finite unions: We know that a set in the reals is compact precisely when it is closed and bounded. So, we solve the problem if we show that closed and bounded sets are closed under the above two operations. First, the arbitrary intersection of closed sets is still closed (we showed this in class - it follows, after taking complements, from the fact that the arbitrary union of open sets is open.); likewise the intersection of bounded sets is still bounded. Second, the finite union of closed sets is still closed: a finite union of bounded sets A i is bounded from above by and likewise for the lower bound. max i sup A i, 2.) Suppose that A is a set. Show that A is compact iff for every collection B of closed sets satisfying: B 1 B N A { } for every finite subset of B, (1) then the infinite intersection of B contains a point of A: A ( B B B) { }. (2) There are a number of ways to show this. The following is not the shortest, but it is straightforward: Assume A is compact, and assume for contradiction that there is some collection B so that the finite intersection property (1) above holds, but the infinite intersection property (3) does not hold. Notice though that if (3) does NOT hold, this means A ( B B B) = { }. (3)

4 4 HW 4 SOLUTIONS which is the same thing as saying that A is contained in the complement of the intersection of all the elements of B. That is, A ( B B B) c B B B c where the last equality here follows from the DeMorgan laws. So, the set of all B c form an open cover of A. But A is compact, so there is a finite subcover, A B c 1 B c 2... B c N = ( N i=1b i ) c In other words, A ( N i=1b i ) = { }, which is a contradiction to our hypothesis (1) on the set B. Conversely, suppose the second property holds, and we want to try to show that A is compact. Assume for contradiction that A is not compact - so there is an open cover with no finite subcover. If you look at the complements of the sets in this open cover and argue in a manner similar to what we have just done, you ll see that this collection of closed sets violates the property we are assuming. 3.) Suppose B 1,..., B N is an open cover of a compact set A. Can i B i = A? Only in the trivial case where A = B i = { }. To see this, note that the union of open sets is open; it follows that the set A would have to be open, and also (since it is compact) bounded and closed. We saw that R and are then the only possibilities (and R is not bounded). 6.) Show that if A is open, then A + B is open: We need to show that any element of A + B has a small interval around it which is also in A + B. Any element of A + B can be written in the form a + b where a A, b B. Since A is open, we get some M N with (a 1, a + 1 ) A. In particular, just adding b to this interval gives M M (a + b 1 M, a + b + 1 M ) A + B. Show that if A and B compact, then A + B is compact: Suppose that z n is a sequence of points in A + B. By definition, we can choose a n A and b n B with z n = a n + b n. By compactness of A, there is a subsequence a nk with limit a A. By compactness of B, there is a subsequence b mj of the b nk with limit b B. It follows that a mj +b mj converges to a + b as j (since the limit of the sum is the sum of the limits), as desired. Give an example where A and B are closed but A + B is not: A = {n + 1 n = 1, 2,... }, n + 1 B = { n n = 1, 2,... }. These sets are closed (they have no limit points). Notice that everything in A is a fraction while B has only integers; hence, everything in A + B is a fraction (so 0 / A + B). However,

5 A + B does contain the sequence of points 1 n + 1, and this sequence converges to the limit point 0. HW 4 SOLUTIONS 5