{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Assignment 5

# Assignment 5 - Assignment#5 Section 2.2 13b Since A is a...

This preview shows pages 1–2. Sign up to view the full content.

Assignment #5 Section 2.2 13b) Since A is a subset of A U B, every subset of A is a subset of A U B Thus P(A) is a subset of P(A U B). Similarly, P(B) is a subset of P(A U B). Thus, P(A) U P(B) is a subset of P(A U B) 13c) Let A={2} and B={1}. The P(A) U P(B)={{},{2}}U{{},{1}} while P(A U B)={{},{1},{2},{1,2}}. In general, if A is a subset of B or B is a subset of A, then P(A U B)=P(A) U P(B) 16b) Let B be contained in the Reals. Then 0 is in B*, so B* is not empty. 16c) Suppose B=B*. Then it is clear that 0 is in B. Next, suppose 0 is already in B then B* = B U {0} = B 16d) 0 is in B*, so (B*)* = B* by (c) 16e) (A U B)* = (A U B) U {0} by definition = A U (B U {0}) = A U B* But we can also write (A U B)* = (B U A) U {0} = B U (A U {0}) = B U A* That is, we have A U B* = B U A*. Note that either way 0 is already in the set. Applying part (c), we have (A U B)* = A* U B* 17b) Grade C. Need to state x belongs to A-C first, as well as x belongs to B-C afterwards. 17c) Grade A 17f) This is a false claim. Section 2.3 1) f) union is {1,2,...,19}, intersection is empty h) Union is [ -(pi), positive infinity) Intersection is [ -(pi), 0] j) Union is the set of all integers Intersection is {0} k) Union is (0,7/3) Intersection is [1/3,2] l) Union is (negative infinity, positive infinity) Intersection is is { } m) Union is R-Z (the set of all real numbers but not include all the integers.) Intersection is { } n) Union is (-infinity, 1) Intersection is (-1, 0] 2)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 4

Assignment 5 - Assignment#5 Section 2.2 13b Since A is a...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online