1. More Functions and Sets
Rosen 1.8

2. Inverse Image
Let f be an invertible function from set A to set B. Let S be a subset of B. We define the inverse image of S to be the subset of A containing all pre-images of all elements of S.
f-1(S) = {aA | f(a) S} S A
f(a) a B

3. Let f be an invertible function from A to B. Let S be a subset of B
Let f be an invertible function from A to B. Let S be a subset of B. Show that f-1(S) = f-1(S)
What do we know?
f must be 1-to-1 and onto S A B b1 a2 b2 a1 f -1
f-1(S)

4. Let f be an invertible function from A to B. Let S be a subset of B
Let f be an invertible function from A to B. Let S be a subset of B. Show that f-1(S) = f-1(S)
Proof: We must show that f-1(S)  f-1(S) and that f-1(S)  f-1(S) .
Let x  f-1(S). Then xA and f(x)  S. Since f(x)  S, x  f-1(S). Therefore x  f-1(S).
Now let x  f-1(S). Then x  f-1(S) which implies that f(x)  S. Therefore f(x)  S and x  f-1(S)
If x  f-1(S). Then f(x)  S after taking the inverse of both sides (i.e., x = f-1(y) means y = f(x); therefore y  S).

5. Let f be an invertible function from A to B. Let S be a subset of B
Let f be an invertible function from A to B. Let S be a subset of B. Show that f-1(S) = f-1(S)
Proof:
f-1(S) = {xA | f(x)  S} Set builder notation
= {xA | f(x)  S} Def of Complement
= f-1(S) Def of Complement

6. Floor and Ceiling Functions
The floor function assigns to the real number x the largest integer that is less than or equal to x. x
x = n iff n  x < n+1, nZ
x = n iff x-1 < n  x, nZ
The ceiling function assigns to the real number x the smallest integer that is greater than or equal to x. x
x = n iff n-1 < x  n, nZ
x = n iff x  n < x+1, nZ

7. Examples 0.5 = 1 0.5 = 0 -0.3 = 0 -0.3 = -1 6 = 6 6 = 6
-3.4 = -3
3.9 = 3

8. Prove that x+m = x + m when m is an integer.
Proof: Assume that x = n, nZ.
Therefore n  x < n+1.
Next we add m to each term in the inequality to get n+m  x+m < n+m+1.
Therefore x+m = n+m = x + m
x = n iff n  x < n+1, nZ

9. Let xR. Show that 2x = x + x+1/2
Proof: Let nZ such that x = n. Therefore
n  x < n+1. We will look at the two cases:
x  n + 1/2 and x < n + 1/2.
Case 1: x  n + 1/2
Then 2n+1  2x < 2n+2, so 2x = 2n+1
Also n+1  x + 1/2 < n+2, so x + 1/2  = n+1
2x = 2n+1 = n + n+1 = x + x+1/2 n
n+1

10. Let xR. Show that 2x = x + x+1/2
Case 2: x < n + 1/2
Then 2n  2x < 2n+1, so 2x = 2n
Also n  x + 1/2 < n+1, so x + 1/2  = n
2x = 2n = n + n = x + x+1/2

11. Characteristic Function
Let S be a subset of a universal set U. The characteristic function fS of S is the function from U to {0,1}such that fS(x) = 1 if xS and fS(x) = 0 if xS.
Example: Let U = Z and S = {2,4,6,8}.
fS(4) = 1
fS(10) = 0

12. Let A and B be sets. Show that for all x, fAB(x) = fA(x)fB(x)
Proof: fAB(x) must equal either 0 or 1.
Suppose that fAB(x) = 1. Then x must be in the intersection of A and B. Since x AB, then xA and xB. Since xA, fA(x)=1 and since xB fB(x) = 1. Therefore fAB = fA(x)fB(x) = 1.
If fAB(x) = 0. Then x  AB. Since x is not in the intersection of A and B, either xA or xB or x is not in either A or B. If xA, then fA(x)=0. If xB, then fB(x) = 0. In either case fAB = fA(x)fB(x) = 0.

13. Let A and B be sets. Show that for all x, fAB(x) = fA(x) + fB(x) - fA(x)fB(x)
Proof: fAB(x) must equal either 0 or 1.
Suppose that fAB(x) = 1. Then xA or xB or x is in both A and B. If x is in one set but not the other, then fA(x) + fB(x) - fA(x)fB(x) = 1+0+(1)(0) = 1. If x is in both A and B, then fA(x) + fB(x) - fA(x)fB(x) = 1+1 – (1)(1) = 1.
If fAB(x) = 0. Then xA and xB. Then fA(x) + fB(x) - fA(x)fB(x) = – (0)(0) = 0.

14. A B AB fAB(x) fA(x) + fB(x) - fA(x)fB(x)
Let A and B be sets. Show that for all x, fAB(x) = fA(x) + fB(x) - fA(x)fB(x)
A B AB fAB(x) fA(x) + fB(x) - fA(x)fB(x)
(1)(1) = 1
(1)(0) = 1
(0)(1) = 1
)-(0)(0) = 0