Let's consider a specific example to illustrate how a non-injective function can cause the equation $f(A \cap B) = f(A) \cap f(B)$ to fail.

Example:

Let $f: \mathbb{R} \to \mathbb{R}$ be the function defined by $f(x) = x^2$. This function is not injective because different inputs can produce the same output. For example, $f(2) = 4$ and $f(-2) = 4$.

Now, let's define two sets $A$ and $B$ as follows:

• $A = \{-2, 1\}$
• $B = \{2, 1\}$

Step 1: Find $A \cap B$

The intersection of $A$ and $B$ is: $A \cap B = \{1\}$

Step 2: Find $f(A \cap B)$

Now, apply the function $f(x) = x^2$ to the intersection: $f(A \cap B) = f(\{1\}) = \{1^2\} = \{1\}$

Step 3: Find $f(A)$ and $f(B)$

Next, find the images of sets $A$ and $B$ under $f$: $f(A) = f(\{-2, 1\}) = \{(-2)^2, 1^2\} = \{4, 1\}$ $f(B) = f(\{2, 1\}) = \{2^2, 1^2\} = \{4, 1\}$

Step 4: Find $f(A) \cap f(B)$

The intersection of $f(A)$ and $f(B)$ is: $f(A) \cap f(B) = \{4, 1\} \cap \{4, 1\} = \{4, 1\}$

Conclusion:

Now, compare $f(A \cap B)$ with $f(A) \cap f(B)$: $f(A \cap B) = \{1\}$ $f(A) \cap f(B) = \{4, 1\}$

Clearly, $f(A \cap B) \neq f(A) \cap f(B)$, showing that the equation fails when $f$ is not injective.

Would you like to explore more about this, or do you have any other questions?

---

Related Questions:

1. How do injective functions affect set operations like union and difference?
2. What are some other examples of non-injective functions?
3. How can you determine if a function is injective from its graph?
4. Can a function be surjective but not injective? Provide an example.
5. How does bijectivity ensure the equality of $f(A \cap B) = f(A) \cap f(B)$?

Tip:

When dealing with functions and set operations, constructing specific examples with small, manageable sets can help clarify how function properties like injectivity affect the outcome.