Can you solve that exercise?---**Exercise 2.** Let $\Omega_1, \Omega_2$ be non-empty sets, $f: \Omega_1 \to \Omega_2$ an arbitrary mapping and $\mathcal{C} \subseteq \mathcal{P}(\Omega_2)$ an arbitrary collection of subsets of $\Omega_2$. Then the following statements hold. (i) If $A, B \subseteq \Omega_2$ and $A \subseteq B$, then $f^{-1}(A) \subseteq f^{-1}(B)$.

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Let's examine Exercise 2, which deals with preimage properties of functions. We have a mapping f from set Omega 1 to set Omega 2. Statement i claims that if A is a subset of B, both subsets of Omega 2, then the preimage of A is a subset of the preimage of B. This diagram illustrates the concept: set A is contained within set B in the codomain, and correspondingly, the preimage of A is contained within the preimage of B in the domain. Now let's outline our proof strategy. To prove that the preimage of A is a subset of the preimage of B, we need to show that every element in the preimage of A is also in the preimage of B. We'll take an arbitrary element x from the preimage of A. By definition of preimage, this means f of x is in A. Since A is a subset of B, f of x must also be in B. Therefore, by definition of preimage again, x must be in the preimage of B. This completes our proof strategy. Here is the formal mathematical proof. We start by letting x be an arbitrary element in the preimage of A. By the definition of preimage, this means f of x is in A. We are given that A is a subset of B. Since f of x is in A, and A is a subset of B, it follows that f of x is also in B. By the definition of preimage again, if f of x is in B, then x is in the preimage of B. Therefore, we have shown that the preimage of A is a subset of the preimage of B, which completes our proof. Let's verify our theorem with a concrete example. Consider domain Omega 1 with elements 1, 2, 3, 4, 5, and codomain Omega 2 with elements a, b, c, d. The function f maps 1 to a, 2 to b, 3 to a, 4 to c, and 5 to c. Let A be the set containing a and b, and B be the set containing a, b, and c. Notice that A is indeed a subset of B. The preimage of A contains elements 1, 2, and 3, while the preimage of B contains all elements 1, 2, 3, 4, and 5. We can clearly see that the preimage of A is a subset of the preimage of B, confirming our theorem. In conclusion, we have successfully proven that statement i of Exercise 2 is true. We demonstrated that if A is a subset of B, both being subsets of the codomain, then the preimage of A is a subset of the preimage of B. This fundamental property shows that preimage operations preserve subset relationships. The proof relied on basic definitions from set theory and holds for any function f. This property is widely used in mathematics, particularly in topology, measure theory, and functional analysis. Exercise 2 part i is now completely solved.