Why is f being an injection equivalent to this ?

In summary, for a function $f$ to be an injection, it must satisfy the property that if $f(A)\cap f(B)\subset f(A\cap B)$ for all sets $A$ and $B$, then $f(A\cap B)$ is nonempty and $A=B$. The proof for this is shown by assuming that $f(A)\cap f(B)\subset f(A\cap B)$ and proving that this implies $f$ is an injection. On the other hand, if $f$ is an injection, then $f(A)\cap f(B)\subset f(A\cap B)$ for all sets $A$ and $B$. This is proven by considering singleton sets $A$ and $
  • #1
fatineouahbi
10
0
1.Prove f(A⋂B) ⊂ f(A) ⋂ f(B)
2.Prove f(A) ⋂ f(B) ⊂ f(A⋂B) ⟺ f is an injection

I've solved the first question , as for the second I started with f(A) ⋂ f(B) ⊂ f(A⋂B) ⇒ f is an injection this way :
Let's suppose f(a) = f(b) = p
If we consider A = {a} and B = {b} then f(A) = f(B) = p
then f(A) ⋂ f(B) = p
then f(A⋂B) = p (because f(A) ⋂ f(B) ⊂ f(A⋂B) from the supposition and f(A⋂B) ⊂ f(A) ⋂ f(B) from the first question)
then A⋂B ≠ ∅
then a=b
then f is an injection .

But I don't know how to solve "f is an injection ⇒ f(A) ⋂ f(B) ⊂ f(A⋂B)" :confused:
 
Physics news on Phys.org
  • #2
$(\implies)$ Assume $f(A)\cap f(B) \subset f(A\cap B)$ for all sets $A$ and $B$. Suppose $A=\{x\}$ and $B=\{y\}$ are singleton sets. Further assume that $f(x)=f(y)$. Then $f(A)=f(B)$, so $f(A)\cap f(B)$ is nonempty. By assumption, $f(A\cap B)$ is nonempty, since $f(A)\cap f(B)\subset f(A\cap B)$. This implies $A\cap B$ is nonempty. Since $A$ and $B$ are singletons, this implies $A=B$, or $\{x\}=\{y\}$, or $x=y$. Hence, $f$ is an injection.

$(\impliedby)$ This is a "what do you know" kind of proof. Can you write it out?
 

Related to Why is f being an injection equivalent to this ?

1. Why is f being an injection equivalent to this?

Being an injection means that for every element in the input set, there is only one corresponding element in the output set. This is equivalent to saying that every element in the output set has a unique pre-image in the input set under the function f.

2. What is the significance of f being an injection?

An injection is significant because it ensures that there is no duplication of elements in the output set, and each element in the output set is uniquely mapped from an element in the input set. This is important in maintaining the integrity and accuracy of the function.

3. How does f being an injection affect the range of the function?

If f is an injection, it means that every element in the output set is mapped from a unique element in the input set. This means that the range of the function is equivalent to the cardinality (number of elements) of the input set, as there are no duplicate elements in the output set.

4. Is f being an injection the same as f being one-to-one?

Yes, f being an injection is equivalent to f being one-to-one. Both terms mean that each element in the input set is mapped to a unique element in the output set. This ensures that there are no duplications or repetitions in the function's mapping.

5. How can we prove that f is an injection?

To prove that f is an injection, we can use the definition of an injection: for every pair of elements x and y in the input set, if f(x) = f(y), then x = y. This means that if two different elements in the input set have the same output, then the elements must be equal. If this condition is met, then f is an injection.

Similar threads

  • Set Theory, Logic, Probability, Statistics
Replies
19
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
5
Views
316
  • Set Theory, Logic, Probability, Statistics
Replies
2
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
544
  • Precalculus Mathematics Homework Help
Replies
4
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
9
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
590
  • Set Theory, Logic, Probability, Statistics
Replies
12
Views
1K
Back
Top