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Images and Inverse Image of a Set

Purpose of Section: To introduce the concept of the image and inverse image
of a set. We show that unions of sets are preserved under a mapping whereas
only one-to-one functions are preserved. On the other hand the inverse
function preserves both unions and intersections.


Parallel light shining on an object creates a shadow behind the object,
possibly falling on a wall, which defines a mapping that sends points on the
object where light strikes to its “shadow” point on a wall; i.e. the point of
continuation of the ray of light on the wall if the ray could pass through the
object. Although we can think of this map as a function sending points on the
object to points on the wall, a more useful way is to think of it as a mapping
between sets; i.e. a set of points on the surface of the object to the shadow
set on the wall.

Often in mathematics, particularly in analysis and topology, one is interested
in finding the set of image points of a function acting on a given set, which
brings us to the following definition.

Definition: Let f : X →Y and . As x moves about the set A , the set
of values f ( x) define the image of A . That is

Also if  we define the inverse image of C as the set

Note that is a well defined set regardless of whether the function f
has an inverse.

Example 1 Let X = {1, 2,3,4} and Y = {a,b,c} and define a function f on
A = {1, 2,3}by f (1) = a, f (2) = a, f (3) = c . Then f ( A) = {a,c} . Also for
example . See Figure 1.

Figure 1

Example 2 Let f : X →Y be given by . See Figure 2.

Figure 2

Then we have

a) f ({−1,1}) = {2} since both f (−1) = f (1) = 2 .

b) f ([−2,2]) = [1,5]

c) f ([−2,3]) = [1,10]

Margin Note: It is often important to know if certain properties of sets are
preserved under certain types of mappings. For instance if X is a connected
set and f a continuous function, then f ( X ) is also connected.

Images of Intersections and Unions

The following theorem gives an important relation of the image of the
intersection of two sets.

Theorem 1

If f : X →Y and , then the images of intersections satisfy

Proof: Let . Hence, there exists an such that
f ( x) = y . Hence x∈ A and x∈B . But

Hence .

Let us now try to show the converse; that is .
Letting we have y∈ f ( A) and y∈ f (B) . Hence

from which we conclude which is only true if the function f is
one-to-one. Hence, we suspect the conclusion is false in general, and thus
we look for a counterexample. That is is a non 1−1 function f X →Y and
subsets and a point that is not in .

To find a counterexample let X = Y =R and A = [−1,0], B = [0,1] . Then
if f is defined by , we have

Hence . See Figure 3.

Counterexample to show

Figure 3

In general, intersections of sets are not preserved under the image of a
function. However, the following theorem shows intersections are preserved.

Theorem 2 Let f : X →Y be a one-to-one function. If , then
intersection is preserved under mappings:

Proof: It suffices to show . Letting ,
we have y∈ f ( A) and y∈ f (B) . Hence

But f is assumed 1-1 so we conclude and so

and and thus

Hence which proves theorem.