Chapter - 5, Functions

Section - 5.4 - Images and Inverse Images: A Research Project


Summary

  • Suppose and . Then the image of under is the set defined as follows:
    .
  • If , then the inverse image of under is the set defined as follows:
    .
  • In this section, we also learned a way to come up with counter examples. It might not help always for finding counter examples but it may atleast give a starting point. The outline is as follows:
    • Try to prove the theorem.
    • If theorem is not correct, we will get stuck at some point.
    • Try to use the information gathered from the point where we are stuck in the proof to come up with counter example.

Soln1

(a)

Yes it will always be true. Proof:

()Suppose . Thus for some , . Thus we have two cases:

Case 1:
Thus and , it follows . Or we can also say .

Case 2:
Thus and , it follows . Or we can also say .

Thus from both cases, . Since is arbitrary, it follows that .

() Suppose . Thus we have two possible cases:

Case 1: .
Thus for some , . Since , it follows . Since and , it follows .

Case 2: .
Thus for some , . Since , it follows . Since and , it follows .

Thus from both cases . Since is arbitrary, it follows that .

Since and , it follows that .

(b)

No, it is not true. Counter example:

Suppose .
.

Thus, .
.
.

Also, , Thus .

Clearly .

(c)

No, it is not true. Counter example:

Note: Earlier solution was wrong. Check the comments.

Updated Solution:

Suppose .
.

Clearly .

Now, and . Clearly but .

Old(Wrong) Solution:

Suppose .
.

Clearly .

Now, and . Clearly . Thus but .


Soln2

(a)

Yes, it will be always true. Proof:

()Suppose . Thus and . Since and , it follows . Similarly, since and , it follows . Thus . Since is arbitrary, it follows .

()Suppose . Thus and . It follows that and and . It follows . Since and , it follows that . Since is arbitrary, it follows .

Thus from both directions, we can conclude

(b)

Yes, it will be always true. Proof:

()Suppose . Thus and f(x) \in Y \lor f(x) \in Z $$. Thus we have two cases:

Case 1:
Since and , it follows .

Case 2:
Since and , it follows .

Thus we have or . It follows .

Since is arbitrary, it follows .

()Suppose . Thus or . It follows we have following cases:

Case 1:
Thus and . Or we can also say .

Case 2:
Thus and . Or we can also say .

It follows from all possible cases that and . Thus . Since is arbitrary, it follows that .

Thus we have and . We can conclude that .

(c)

Yes, it will be always true. Proof:

()Suppose . Thus and . Thus . Since and , it follows . Also, since and , it follows . Since and , it follows that . Since is arbitrary, it follows .

()Suppose . Thus and . Since , it follows and . Since and , it follows . Thus we have and . It follows . Since and , it follows that . Since is arbitrary, it follows that .

Now we can conclude from both directions that .

(d)

No, it is not always true. Counter example:

Suppose .

Thus .

Also, and . Thus .

Thus even if , .


Soln3 False.

Suppose .

Thus and . Thus . Thus if is not one-to-one, theorem is not correct.


Soln4 False.

Suppose .

Also, and . Thus . Thus if is not onto, theorem is incorrect.


Soln5 TODO


Soln6

Suppose denotes inverse image of under . Thus .

Also suppose denotes image of under . Thus .

()Suppose . Thus and . Suppose . Thus and . Thus there exists an element such that . It follows that . Thus .

()Suppose . Thus for some element we have . Since is one-to-one and onto, and , it follows . Since , it follows . Since and , it follows . Thus .

Since and , we can conclude that .