Math 3000 Homework Answers #27

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. 300 : 5, 7, 9, 10, 13, 15, 17, 18, 20 )

5.(a) open
(c) neither open nor closed
(d) neither open nor closed
(f) closed
(g) open
(h) open
(j) neither open nor closed.
7. (a) (c) (d) (e)The complement of AB is the intersection of the complements of A and B. Since A and B are closed sets, their complements are open sets. Since the intersection of any finite number (2 in this case) of open sets is open, this complement is open. Therefore, AB is closed.
(f)
9. An interior point of a set is a point which is contained in some open interval which is contained in the set. Note that every point in an open interval is an interior point. For each interior point of set A, pick an open interval containing that point and contained in A. The union of these open intervals contains all the interior points of A. Any point in this union is in an open interval of A and so is an interior point. Thus, the union consists of exactly the interior points of A. Since it is a union of open sets, it is an open set. Therefore, the set of interior points of A is an open set.
10. (a) The boundary points of (2, 5] are {2, 5}. The boundary points of (0 , 1) are {0, 1}. The boundary points of [3, 5]{6} are {3, 5, 6}. All the points of are boundary points.
(b) Suppose that x is a boundary point of A. Then every neighborhood of x contains points that are in A and points that are not in A. Thus, no neighborhood of x contains only points of A, so x is not an interior point of A. Also, no neighborhood of x contains points only in the complement of A, so x is not an interior point of the complement of A.
Now suppose, that x is not an interior point of A nor of the complement of A. Take any neighborhood of x. Since x is not an interior point, this neighborhood can not consist only of points of A, so there must exist some point in the complement of A in this neighborhood. Since x is not an interior point of the complement of A, there must exist some point in this neighborhood which is in A. Therefore, every neighborhood of x contains points points of A and points not in A, so x is a boundary point of A.
(c) If A is open, then every point of A is an interior point of A. Since, by (b), a boundary point can not be an interior point of A, no boundary point is contained in A. On the other hand, if A contains no boundary points then, if A is empty it is an open set, otherwise it contains a point. Since this point is not a boundary point, some neighborhood of it does not contain any points of the complement of A, i.e., all of its points are in A and so, it is an interior point of A. Since this is true for all the points of A, A is open.
(d) If A is closed, then the complement of A is open. By (c), the complement of A contains no boundary point, so, all the boundary points of A must be contained in A. On the other hand, if A contains all of its boundary points, none of its boundary points are contained in the complement of A. This means, by (c), that the complement of A is open, so A is closed.
13. (a) Consider any open cover of AB. Since A is a subset of AB, this cover is also an open cover of A. Since A is compact, there exists a finite subcover which covers A, call it F. Since B is a subset of AB, and B is compact, there exists a finite subcover which covers B, call it F'. Now, FF' is a finite subcover which covers both A and B, so it covers AB. Therefore, AB is compact.
(b) Let O be an open cover of AB. Since A and B are compact, they are closed and so is AB. Let C be the complement of AB. C is an open set and O together with C is an open cover of A (actually of any subset at all since it is an open cover of ). Since A is compact, this open cover has a finite subcover. By removing C from this subcover we get a finite subcover of O which covers AB, therefore AB is compact.
15. (b) compact
(c) compact
(e) compact
(f) compact
(h) compact
17. (a) The sets in C are open intervals, so we need only show that they cover (0,1]. The right hand endpoints of the intervals in C are always greater than 1, so, to show that this is a cover we must show that for any x in (0,1] we can find a natural number n so that (n+2)/2n < x. Since x is positive, and the limit of (n+2)/2n as n goes to infinity is 0, there will always exist an n so that (n+2)/2n < x. This shows that C is a cover.
(b) There is no finite subcover of C that covers (0,1], since such a finite cover would contain an interval with a smallest left hand endpoint and any positive number less than this endpoint would not be covered by the finite cover.
(c) Since (0,1] is clearly bounded, the Heine-Borel theorem says that it is not closed (otherwise it would be compact).

18. Consider the intersection of an arbitrary number of compact sets. This intersection is contained in any one of them. Since each set is compact, the Heine-Borel theorem says that each set is closed and bounded. The intersection, being a subset of a bounded set is itself bounded. The intersection of an arbitrary number of closed sets is closed. Therefore, by the Heine-Borel theorem the intersection is compact, since it is closed and bounded.

20. (a) C. The chosen delta is negative, it should be x - a instead.
(b) C It must be shown that every open cover of AB has a finite subcover. A cover of just A, may not be a cover of AB, and similarly for B. However, any open cover of AB is an open cover of both A and B, so the proof should start with an open cover of AB rather than seperate covers for A and B.
(d) A.