Proofs involving functions.

1
Proofs involving functions.
Let f A?B be a function. Then for any subset of
A, C?A, we can define the set of images for
elements of C that we denote as f (C) f (C)
f (x) x?C .
2
For any subset of B, D?B, we can define the set
of pre-images of elements from D, which we
denote as f -1(D) f -1(D) x x?A and f
(x)?D .
Pay attention that f -1(D) is just a notation for
the set of pre-images, which does not assumes
the existence of inverse function for f.
3
Example A a, b, c, B x, y, z, and f
(a)x, f (b)y, f (c)y.
In this example, function f is neither injective
nor surjective. So, inverse function f -1 does
not exist (in other words, inverse relation of f
is not a function). But we can find a set of
pre-images for any subset of B. For example f
-1(z)? f -1(x, z)a f -1(x, y)
a, b, c.
4
Questions

• Is it always true, that f (A) B?

Ans. It is true only if f is surjective.
Otherwise f (A)?B.
5

• Is it always true that f -1(B)A?

Ans. Yes, since f is a function, any element of A
has an image that belongs B. As a result any
element of A appears as a pre-image of some
element of B.
6

• Is it true that if f(C)f(D), where C, D?A, then
  CD?

C
f(D)
D
Ans. This is true only if f is injective.
7

• Is it true that if f -1(C)f -1(D), where C, D ?
  B, then CD?

C
f -1(C)
f -1(D)
Ans. This is true only if f is surjective
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Example. Prove that if f is injective, then X?Y?
implies f(X)?f(Y)?.
f
B
A
X
f (X)
Y
f (Y)
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Proof by contradiction. Assume that X?Y? and
f(X)?f(Y)??, (1).
(1) implies that there exists some common
element, z?f(X) and z? f(Y), (2).
From the definition of the set f(X) and f (Y) (2)
implies that there exists x?X, such that f(x)z
(3) and there exists y?Y, such that f(y)z
(4).
So, we have f(x) f(y)z, (5) and since f is
injective (5) implies that xy, (6).
Since x?X and y?Y , (6) implies that X?Y?? in
contradiction with assumption. The
contradiction proves the initial statement.
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Let f A ? B be a function. Prove or disprove
that for any D ? B, f (f ?1(D )) D
f ?1(D)
f (f ?1(D ))
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f ?1(D)
f (f ?1(D )) ?D
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Suppose R ?A ? A is an equivalence relation.
Prove or disprove that if R is a function, then
this function is IA (identity function on A)
Proof by contradiction. Assume that R is an
equivalence relation on A that defines a
function f A ?A , and f ?IA .
It means that there exist a ?b , such that f
(a)b. At the same time (a, b) ?R .
Since R is an equivalence relation, it is
reflexive, symmetric and transitive. It implies,
that (b, a) ?R (symmetric).
By the transitive property (a, b) ?R and (b, a)
?R imply (a, a) ?R
But (a, b) ?R and (a, a) ?R means that R is not a
function!
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2, p. 277
a) If f A ?B and (a, b), (a, c ) ? f , then b
c
b) If f A ?B is a one-to one correspondence
(injective) and A and B are finite, then AB
c) If f A ?B is a one-to one (injective), then
f is invertible ( f ?1 is a function).
d) If f A ?B is invertible ( f ?1 is a
function), then f is one-to one (injective).
e) If f A ?B is a one-to one (injective) and g,
h B ?C with g ? f h ? f , then g h .
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f) If f A ?B and A1, A2 ? A, then f (A1? A2 )
f (A1) ? f (A2 )
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g) If f A ?B and B1, B2 ? B, then f
?1(B1? B2 ) f ?1(B1) ? f ?1(B2 )
6, p. 277. Let A 1, 2, 3, 4, 5 and B 1,
2, 3, 4, 5, 6. How many injective functions f
A ?B satisfy a) f (1) 3 ?
b) f (1) 3, f (2) 6 ?
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Let R ? A?A be a transitive relation. Prove or
disprove that s(R) is transitive.
Let R ? A?A be an equivalence relation.
i)Prove that R?R is equivalence relation.
ii)Prove or disprove that R and R?R induce the
same partition of A.
Let f ?A?B and g ?A?B are two relations.
Suppose the relations f, g and f ?g are
functions from A to B (That is for any
element a?A there exists a unique element b?B,
(a, b) ? f . The same for g and f ?g.) Prove
that f g.
    Let A 1, 2, 3, 4, 5? 1, 2, 3, 4,
5, and define R on A by (x1, y1)R(x2,
y2) if x1 y1 x2y2. i) verify that R is
equivalence relation. ii) Determine the
equivalence classes (1, 3), (2, 4), and
(1,1). iii) Determine the partition of A
induced by R.