Math Review

Solutions to Homework #2

  1. (a) f -1(f(A0))  A0.

Proof: Choose any a  A0. By definition of f(A0), it follows that f(a)  f(A0). By definition of f -1, we have f -1(C) = {a  A: f(a)  C}, C  B. Letting C = f(A0) gives f -1(f(A0)) = {a  A: f(a)  f(A0)}. This implies that a  f -1(f(A0)).

Thus, f -1(f(A0))  A0.

(b) f(f -1(B0))  B0.

Proof: Choose any b  f(f -1(B0)). By definition of f, it follows that there exists an a  f -1(B0) where b = f(a). By definition of f -1(B0), this implies that b = f(a)  B0.

Thus, f(f -1(B0))  B0.

  1. (a) A(BC) = (AB)(AC).

Proof: Let x  A(BC). It follows that x  A and x  (BC). This implies that (x  AB) or (x  AC). This can be written as x  (AB)(AC). It follows that A(BC)  (AB)(AC).

Now, let y  (AB)(AC). It follows that y  (AB) or y  (AC). This implies that (y  A) and (y  BC). This can be written as y  A(BC). It follows that (AB)(AC)  A(BC).

Combining these two results yields A(BC) = (AB)(AC).

(b) A(BC) = (AB)(AC).

Proof: Let x  A(BC). It follows that x  A or x  (BC). This implies that (x  AB) and (x  AC). This can be written as x  (AB)(AC). It follows that A(BC)  (AB)(AC).

Now, let y  (AB)(AC). It follows that y  (AB) and y  (AC). This implies that (y  A) or (y  BC). This can be written as y  A(BC). It follows that (AB)(AC)  A(BC).

Combining these two results yields A(BC) = (AB)(AC).

3.(a) not open, not closed.

(b) closed

(c) closed

(d) open

(e) closed

4.(a) not open, closed, not compact.

(b) not open, closed, compact.

(c) not open, closed, not compact.

(d) not open, not closed, not compact.

(e) not open, closed, not compact.

5.A closed subset of a compact set is closed.

Proof: Assume that we are considering subsets of Rn. Suppose A  B, where A is closed and B is compact. Since B is compact, it must be bounded. Thus, there is an open ball of finite radius that contains B. Since AB, it is also contained in this open ball and thus A is bounded. Since A is closed and bounded, it is compact.

6.Every finite set is compact.

Proof: Assume that we are considering subsets of Rn. Let A be a finite set with n elements. Then A can be expressed as the union of n sets, each containing a single element: A = iI{xi}, where I = {1, 2, ..., n}. Since each of these singleton sets is closed, A is also closed since a finite union of closed sets is closed. For any two elements in A, there is a finite distance, d(xi, xj). Let d* denote the largest of these distances. For any finite number t, r = d*+t is finite, since it is the sum of two finite numbers. Thus, we can find an open ball with radius r/2 that contains A. So A is both closed and bounded, and thus compact.

7.Assume that we are considering subsets of Rn.

(a) Any intersection of compact sets is compact.

Proof: Let C = iICi be the intersection of compact sets Ci, where I  N. Each Ci is closed, so C is closed (since any intersection of closed sets is closed). Each Ci is also bounded, and is thus contained in an open ball of finite radius. Each of these open balls also contains C, since C  Ci for all iI. So C is both closed and bounded, and therefore compact.

(b) Any finite union of compact sets is compact.

Proof: Let C = iICi be the finite union of compact sets Ci, where I = {1, 2, ..., n}. Each Ci is closed, so C is closed (since a finite union of closed sets is closed). Choose one point in each set Ci, call it xi. Let d* be the largest distance d(xi, xj) between any two of the selected points. Because the sets containing the two points are bounded, an open ball of finite radius containing each can be found. Choose two such open balls and let r be the radius of the largest. Then there is an open ball with radius d*+4r that contains C. So C is both closed and bounded, and therefore compact.

  1. Let r > 0. Rn can be written as Rn = x [Br(x): x  Rn], where Br(x) is an open ball about x with radius r. Since any arbitrary unions of open sets are open, it follows that Rn is open.

Also, the complement of Rn is empty. Since an empty set is open, it follows that Rn is closed.

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