# Proving Equivalence of 1-1 Function Statements

• Incand
In summary, the given statements a), b), and c) are equivalent, where a) states that a given function f is 1-1, b) states that the image of the intersection of two sets is equal to the intersection of the images of the two sets under f, and c) states that the inverse image of the image of a set is equal to the set itself, for all subsets of S. The proofs for a) ⇒ b) and a) ⇒ c) have been shown, and the remaining equivalences can be proven by assuming the contrapositive and using the given definitions and properties.
Incand

## Homework Statement

Let ##f:S\to T## be a given function. Show the following statements are equivalent:
a) ##f## is 1-1
b) ##f(A\cap B) = f(A) \cap f(B),\; \forall A,B \in S##
c) ##f^{-1}(f(A)) = A,\; \forall A \subseteq S.##

## Homework Equations

Definition:
##f## is 1-1 of ##A## into ##B## provided that ##f(x_1) \ne f(x_2)## whenever ##x_1 \ne x_2, \; \; \; x_1,x_2 \in A##.

Definitions:
Let ##f## is a mapping ##f:A \to B##:
If ##E \subseteq A## then ##f(E)## is the set of all elements ##f(x)## with ##x \in E##.
If ##E \subseteq B## then ##f^{-1}(E)## denotes the set of all ##x\in A## such that ##f(x) \in E##.

## The Attempt at a Solution

I think I'm able to prove a) ##\Longrightarrow## b) and a) ##\Longrightarrow## c) but I can't complete the rest.
Lets first prove the general statement ##A \subseteq f^{-1}(f(A))## :
Take ##\alpha \in A## then ##f(\alpha) \in f(A)## and hence ##\alpha \in f^{-1}(f(A))##.

We can also prove that ##f(A \cap B) \subseteq f(A) \cap f(B)##:
Take ##\alpha \in f(A \cap B)## that means ##\alpha = f(z)## for some ##z\in A \cap B## and hence ##\alpha \in f(A)\cap f(B)##.

It's left to prove the equivalence between
a) ##f## is 1-1
b) ## f(A) \cap f(B) \subseteq f(A\cap B),\; \forall A,B \in S##
c) ##f^{-1}(f(A)) \subseteq A,\; \forall A \subseteq S.##
a) ##\Longrightarrow## b)
Take ##\alpha \in f(A) \cap f(B)## then ##\alpha = f(z_1), \; z_1 \in A## and ##\alpha = f(z_2), \; z_2 \in B##. But since ##f## is 1-1 ##z_1 = z_2## hence ##\alpha \in f(A \cap B)## and ## f(A) \cap f(B) \subseteq f(A\cap B)##.

a) ##\Longrightarrow## c)
Take ##\alpha \in f^{-1}(f(A))## that is ##z = f(\alpha)## for some ##z\in B##. That is
##f(\alpha) \in f(A)## hence ##f(\beta) = z## for some ##\beta \in A## but since ##f## is 1-1 this means ##\alpha = \beta## and ##\beta \in A## so ##f^{-1}(f(A)) \subseteq A##.

To complete the proof I need to either show that c) ##\Longrightarrow## a) and b) ##\Longrightarrow## c) OR show that c) ##\Longrightarrow## a) and b) ##\Longrightarrow## a).

c) ##\Longrightarrow## a)
It's equivalent to show the contrapositive that ##f(x_1) = f(x_2) \Longrightarrow x_1 = x_2##. Take ##x_1, x_ 2 \in A## so that ##f(x_1)= f(x_2)## then by c) ##x_1,x_2 \in f^{-1}(f(A))##. This means that ##z_1 = f(x_1)## and ##z_2 = f(z_2)## for ##z_1,z_2 \in B## but from the premise ##z_1 = z_2##.

I don't seem to get anywhere with the last part nor any luck with any of the other equivalences. Any hints on how to go about it? I'm also wondering If what I've done so far is correct?

Incand said:

## Homework Statement

Let ##f:S\to T## be a given function. Show the following statements are equivalent:
a) ##f## is 1-1
b) ##f(A\cap B) = f(A) \cap f(B),\; \forall A,B \in S##
c) ##f^{-1}(f(A)) = A,\; \forall A \subseteq S.##

## Homework Equations

Definition:
##f## is 1-1 of ##A## into ##B## provided that ##f(x_1) \ne f(x_2)## whenever ##x_1 \ne x_2, \; \; \; x_1,x_2 \in A##.

Definitions:
Let ##f## is a mapping ##f:A \to B##:
If ##E \subseteq A## then ##f(E)## is the set of all elements ##f(x)## with ##x \in E##.
If ##E \subseteq B## then ##f^{-1}(E)## denotes the set of all ##x\in A## such that ##f(x) \in E##.

## The Attempt at a Solution

I think I'm able to prove a) ##\Longrightarrow## b) and a) ##\Longrightarrow## c) but I can't complete the rest.
Lets first prove the general statement ##A \subseteq f^{-1}(f(A))## :
Take ##\alpha \in A## then ##f(\alpha) \in f(A)## and hence ##\alpha \in f^{-1}(f(A))##.

We can also prove that ##f(A \cap B) \subseteq f(A) \cap f(B)##:
Take ##\alpha \in f(A \cap B)## that means ##\alpha = f(z)## for some ##z\in A \cap B## and hence ##\alpha \in f(A)\cap f(B)##.

It's left to prove the equivalence between
a) ##f## is 1-1
b) ## f(A) \cap f(B) \subseteq f(A\cap B),\; \forall A,B \in S##
c) ##f^{-1}(f(A)) \subseteq A,\; \forall A \subseteq S.##
a) ##\Longrightarrow## b)
Take ##\alpha \in f(A) \cap f(B)## then ##\alpha = f(z_1), \; z_1 \in A## and ##\alpha = f(z_2), \; z_2 \in B##. But since ##f## is 1-1 ##z_1 = z_2## hence ##\alpha \in f(A \cap B)## and ## f(A) \cap f(B) \subseteq f(A\cap B)##.

a) ##\Longrightarrow## c)
Take ##\alpha \in f^{-1}(f(A))## that is ##z = f(\alpha)## for some ##z\in B##. That is
##f(\alpha) \in f(A)## hence ##f(\beta) = z## for some ##\beta \in A## but since ##f## is 1-1 this means ##\alpha = \beta## and ##\beta \in A## so ##f^{-1}(f(A)) \subseteq A##.

To complete the proof I need to either show that c) ##\Longrightarrow## a) and b) ##\Longrightarrow## c) OR show that c) ##\Longrightarrow## a) and b) ##\Longrightarrow## a).

c) ##\Longrightarrow## a)
It's equivalent to show the contrapositive that ##f(x_1) = f(x_2) \Longrightarrow x_1 = x_2##. Take ##x_1, x_ 2 \in A## so that ##f(x_1)= f(x_2)## then by c) ##x_1,x_2 \in f^{-1}(f(A))##. This means that ##z_1 = f(x_1)## and ##z_2 = f(z_2)## for ##z_1,z_2 \in B## but from the premise ##z_1 = z_2##.

I don't seem to get anywhere with the last part nor any luck with any of the other equivalences. Any hints on how to go about it? I'm also wondering If what I've done so far is correct?
a) ⇒ b) and a) ⇒ c) are correct.
I don't exactly understand what you did in c) ⇒ a)

Hint for b) ⇒ c)
Take ##A \subset S##, and set ##B=f^{-1}(f(A)) \setminus A##. Use b) to prove that ##B= \varnothing##.

Hint for c) ⇒ a)
Take ##x\in S## and apply c) to ##A=\{x\}##.

Incand
Cheers! The hints really helped! Think I got them now.

b) ##\Longrightarrow## c)
Let ##A \subseteq S## and set ##B = f^{-1}(f(A))\backslash A## then ##A \cap B = \varnothing##. Using b)
##f(\varnothing ) = f(A \cap B) = f(A)\cap f(B), \; \; \forall A \subseteq S##. Hence
##f(\varnothing ) =f(B)## but ##f(\varnothing) = \varnothing## and ##f(B) = \varnothing## only when ##B = \varnothing## so ##B=\varnothing##.
This gives us that ##f^{-1}(f(A)) = A, \; \; \forall A\subseteq S##.

c) ##\Longrightarrow## a)
Take ##x\in S## and take ##A = \{x\}## then by c) ##f^{-1}(f(A)) = A= \{x\}##. Since the inverse image of ##f(A)## has only one element there is only one ##x## satisfying ##z=f(x)## for each ##x\in S##. That means if ##x_1 \ne x_2## ##f(x_1) \ne f(x_2)## and hence ##f## is 1-1.

Samy_A

## 1) What is a "range of mappings proof"?

A range of mappings proof is a type of mathematical proof that demonstrates that a specific set of values can be mapped to a specific set of outputs within a given function or system.

## 2) How is a range of mappings proof different from other types of mathematical proofs?

A range of mappings proof specifically focuses on the outputs of a function or system, rather than the inputs or the overall behavior of the function or system. It is often used to prove the existence or non-existence of certain values within a given range.

## 3) What is the purpose of a range of mappings proof?

The purpose of a range of mappings proof is to provide evidence or justification for the outputs of a function or system within a specific range. It can help to verify the accuracy or validity of a given mathematical model or theory.

## 4) What are some common techniques used in range of mappings proofs?

Some common techniques used in range of mappings proofs include mathematical induction, proof by contradiction, and the use of logical equivalences. These techniques can help to systematically demonstrate the relationship between inputs and outputs within a given range.

## 5) How can a range of mappings proof be applied in real-world situations?

A range of mappings proof can be applied in a wide range of fields, such as physics, economics, and computer science. It can be used to analyze and predict the behavior of systems or processes, and to validate the accuracy of mathematical models. For example, a range of mappings proof could be used to demonstrate the range of possible outcomes in a stock market model or to prove the existence of a solution to a physics problem.

• Calculus and Beyond Homework Help
Replies
1
Views
723
• Calculus and Beyond Homework Help
Replies
9
Views
768
• Calculus and Beyond Homework Help
Replies
2
Views
2K
• Calculus and Beyond Homework Help
Replies
2
Views
2K
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
4
Views
2K
• Calculus and Beyond Homework Help
Replies
8
Views
2K
• Calculus and Beyond Homework Help
Replies
6
Views
448
• Calculus and Beyond Homework Help
Replies
8
Views
404
• Calculus and Beyond Homework Help
Replies
8
Views
842