Math 3000 Homework Answers #28

From Smith, Eggen, & St. Andre, A Transition to Advanced Mathematics, 5th Ed.

Answers to * problems are given in the back of the book and will not be reproduced here.

(pg. 305 : 1, 2, 4, 6, 7, 10, 13, 15 )

1. (a) .
(b) {1/n | n }
(c) {±1 + 1/n | n }
(d) {n + 1/m | n, m }
(e) any interval of real numbers.
2.(b) The derived set is empty (there are no accumulation points).
(d) The derived set consists of all non-negative reals.
(e) The derived set is the closed interval [0,1].
(f) The derived set is the closed interval [3, 7].
(h) The derived set is empty.
(i) The derived set is the set of all reals.
(j) The derived set consists of {0}
4. Let A be a subset of reals, and let z = sup(A). Let N be any neighborhood of z. If b N and b < z, then b is not an upper bound of A. This means that there are elements of A which are greater than b. Since z is an upper bound, these elements are less than or equal to z. By assumption, z is not in A, so these elements can not equal z, i.e., they are in the interval (b, z) which is contained in N. Thus, every neighborhood of z contains points of A (distinct from z), so z is an accumulation point.

6. Since A AB, A' (AB)', by exercise 4(a). Similarly, B' (AB)'. Therefore, A'B' (AB)'.
Now, suppose that x is an accumulation point of AB. Suppose that x is not an accumulation point of A nor an accumulation point of B. Since x is not an accumulation point of A, there is an open interval containing x which does not contain any point of A, call it N. Since x is not an accumulation point of B, there is an open interval containing x which does not contain any point of B, call it N'. Now, NN' is an open interval which is not empty (since it contains x) and so, it contains points other than x. None of these points is in A or B, and this contradicts the fact that x is an accumulation point of AB. Thus, x must be an accumulation point of A or of B, so, we have (AB)' A'B' , proving equality.

7. Let x be an accumulation point of AB. Take any neighborhood of x. It must contain points of AB other than x. Such points are in A, so x is an accumulation point of A. These points are also in B, so x is an accumulation point of B. Thus, x is in the intersection of A' and B'.
Consider the sets A = [0, 1) and B = (1, 2]. AB is empty, so it has no accumulation points. But, A' = [0, 1] and B' = [1, 2], so their intersection is not empty and consists of {1}.
10. (a) Let x be a point in the complement of c(A). Since x is not in A and it is not an accumulation point of A, there exists an open neighborhood about x which contains no point of A (because x is not an accumulation point of A) and no point of A' (if there was an accumulation point of A in this neighborhood, there would have to be a point of A in the neighborhood different from x, and we know that there is no point of A in this neighborhood). So, this neighborhood is contained in the complement of c(A). Therefore, c(A)' s complement is an open set, so c(A) is a closed set.
(b) This is obvious from the definition of c(A).
(d) Since A c(A), A' c(A)'. Since c(A) is closed, c(A)' c(A), by Thm. 7.8. Consider a point x in A which is not an accumulation point of A. There exists an open neighborhood about x which contains no other point of A, call it N. If x is an accumulation point of c(A) then N must contain a point of A' other than x. Let y be a point of A' in N. The only point of A which is in N is x, so every neighborhood of y when intersected with N would give a neighborhood of y which must contain x. For subsets of the reals this is impossible, because you can find a neighborhood of y which does not contain x and then intersect it with N to get a neighborhood of y which contains no element of A. Contradicting the fact that y is an accumulation point of A. (In general however, this argument does not work).
13. If a subset of the real numbers is bounded and has no accumulation point, then by the Bolzano-Weierstrass theorem it can not be infinite. Therefore, the set is finite.

15. (a) F Part ii) of the proof is in error. It claims that (P Q)(P)(Q).
(b) F. The set is a counterexample.
(d) A.
(e)F. The statement is false, consider with subset . The "proof" tries to use the converse of the Bolzano-Weierstrass theorem which is false.