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More Functions and Sets Rosen 1.8 B A S f(a) a 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} B f f A -1

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inverse image

B

A

S

f(a)

a

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}
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

B

f

f

A

-1

a2

b1

b2

a1

S

f-1(S)

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

let f be an invertible function from a to b let s be a subset of b show that f 1 s f 1 s4
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)

let f be an invertible function from a to b let s be a subset of b show that f 1 s f 1 s5
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

floor and ceiling functions
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
examples
Examples

0.5 = 1

0.5 = 0

-0.3 = 0

-0.3 = -1

6 = 6

6 = 6

-3.4 = -3

3.9 = 3

prove that x m x m when m is an integer
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

let x r show that 2x x x 1 2

n

n+1

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

let x r show that 2x x x 1 210
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

characteristic function
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

let a and b be sets show that for all x f a b x f a x f b x
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.

let a and b be sets show that for all x f a b x f a x f b x f a x f b x
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)(0) = 0.

let a and b be sets show that for all x f a b x f a x f b x f a x f b x14
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 1+1-(1)(1) = 1

1 0 1 1 1+0-(1)(0) = 1

0 1 1 1 0+1-(0)(1) = 1

0 0 0 0 0+)-(0)(0) = 0