To prove that f^−1(G ∩ H) = f^−1(G) ∩ f^−1(H), we will use the definition of the inverse image and the properties of sets.

Step 1: Let's start by taking an arbitrary element x from f^−1(G ∩ H). By definition of the inverse image, this means that f(x) is in G ∩ H. By definition of intersection, this means that f(x) is in G and f(x) is in H.

Step 2: Now, since f(x) is in G, this means that x is in f^−1(G) by definition of the inverse image. Similarly, since f(x) is in H, this means that x is in f^−1(H).

Step 3: Therefore, since x is in both f^−1(G) and f^−1(H), this means that x is in f^−1(G) ∩ f^−1(H) by definition of intersection. Since x was an arbitrary element from f^−1(G ∩ H), this shows that f^−1(G ∩ H) is a subset of f^−1(G) ∩ f^−1(H).

Step 4: Now, let's take an arbitrary element x from f^−1(G) ∩ f^−1(H). By definition of intersection, this means that x is in f^−1(G) and x is in f^−1(H). By definition of the inverse image, this means that f(x) is in G and f(x) is in H.

Step 5: Therefore, since f(x) is in both G and H, this means that f(x) is in G ∩ H by definition of intersection. Since x was an arbitrary element from f^−1(G) ∩ f^−1(H), this shows that f^−1(G) ∩ f^−1(H) is a subset of f^−1(G ∩ H).

Step 6: Since we have shown that each set is a subset of the other, this means that the two sets are equal. Therefore, f^−1(G ∩ H) = f^−1(G) ∩ f^−1(H).

Question

To prove that f^−1(G ∩ H) = f^−1(G) ∩ f^−1(H), we will use the definition of the inverse image and the properties of sets.

Step 1: Let's start by taking an arbitrary element x from f^−1(G ∩ H). By definition of the inverse image, this means that f(x) is in G ∩ H. By definition of intersection, this means that f(x) is in G and f(x) is in H.

Step 2: Now, since f(x) is in G, this means that x is in f^−1(G) by definition of the inverse image. Similarly, since f(x) is in H, this means that x is in f^−1(H).

Step 3: Therefore, since x is in both f^−1(G) and f^−1(H), this means that x is in f^−1(G) ∩ f^−1(H) by definition of intersection. Since x was an arbitrary element from f^−1(G ∩ H), this shows that f^−1(G ∩ H) is a subset of f^−1(G) ∩ f^−1(H).

Step 4: Now, let's take an arbitrary element x from f^−1(G) ∩ f^−1(H). By definition of intersection, this means that x is in f^−1(G) and x is in f^−1(H). By definition of the inverse image, this means that f(x) is in G and f(x) is in H.

Step 5: Therefore, since f(x) is in both G and H, this means that f(x) is in G ∩ H by definition of intersection. Since x was an arbitrary element from f^−1(G) ∩ f^−1(H), this shows that f^−1(G) ∩ f^−1(H) is a subset of f^−1(G ∩ H).

Step 6: Since we have shown that each set is a subset of the other, this means that the two sets are equal. Therefore, f^−1(G ∩ H) = f^−1(G) ∩ f^−1(H).

Knowee AI · Accepted Answer

To prove that f^−1(G ∩ H) = f^−1(G) ∩ f^−1(H), we will use the definition of the inverse image and the properties of sets.

Step 1: Let's start by taking an arbitrary element x from f^−1(G ∩ H). By definition of the inverse image, this means that f(x) is in G ∩ H. By definition of intersection, this means that f(x) is in G and f(x) is in H.

Step 2: Now, since f(x) is in G, this means that x is in f^−1(G) by definition of the inverse image. Similarly, since f(x) is in H, this means that x is in f^−1(H).

Step 3: Therefore, since x is in both f^−1(G) and f^−1(H), this means that x is in f^−1(G) ∩ f^−1(H) by definition of intersection. Since x was an arbitrary element from f^−1(G ∩ H), this shows that f^−1(G ∩ H) is a subset of f^−1(G) ∩ f^−1(H).

Step 4: Now, let's take an arbitrary element x from f^−1(G) ∩ f^−1(H). By definition of intersection, this means that x is in f^−1(G) and x is in f^−1(H). By definition of the inverse image, this means that f(x) is in G and f(x) is in H.

Step 5: Therefore, since f(x) is in both G and H, this means that f(x) is in G ∩ H by definition of intersection. Since x was an arbitrary element from f^−1(G) ∩ f^−1(H), this shows that f^−1(G) ∩ f^−1(H) is a subset of f^−1(G ∩ H).

Step 6: Since we have shown that each set is a subset of the other, this means that the two sets are equal. Therefore, f^−1(G ∩ H) = f^−1(G) ∩ f^−1(H).

Prove thatf −1(G ∩ H) = f −1(G) ∩ f −1(H)

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