Let f:A->B, where A and B are nonempty, and let T1 and T2 be subsets of B.
a.Prove that f^-1(T1 U T2)= f^-1(T1) U f^-1(T2).
b.Prove that f^-1(T1 intersects T2) = f^-1(T1) intersects f^-1(T2). I think once I see a I can do b.
c. Prove that f^-1(T1) - f^-1(T2) = f^-1(T1-T2).
d. Prove that if T1 is a proper subset of T2, then f^-1(T1) is a proper subset of f^-1(T2).
I am confused on this one and need some help understanding inverse mapping. Thank you
To prove these statements involving inverse mappings, we will use the definitions and properties associated with inverse images or pre-images.
a. To prove that f^(-1)(T1 U T2) = f^(-1)(T1) U f^(-1)(T2), we need to show that an element x belongs to one side if and only if it belongs to the other side.
Proof:
Let x be an element in f^(-1)(T1 U T2). By definition, this means that f(x) belongs to T1 U T2.
Case 1: f(x) belongs to T1. Then by definition, x belongs to f^(-1)(T1).
Case 2: f(x) belongs to T2. Similarly, this implies x belongs to f^(-1)(T2).
Therefore, in both cases, x belongs to either f^(-1)(T1) or f^(-1)(T2). Hence, x belongs to f^(-1)(T1) U f^(-1)(T2).
Now, let x be an element in f^(-1)(T1) U f^(-1)(T2). This means that x belongs to either f^(-1)(T1) or f^(-1)(T2).
Case 1: x belongs to f^(-1)(T1). By definition, f(x) belongs to T1 U T2.
Case 2: x belongs to f^(-1)(T2). Similarly, f(x) belongs to T1 U T2.
Again, in both cases, f(x) belongs to T1 U T2. Therefore, x belongs to f^(-1)(T1 U T2).
Since we have shown that x belongs to f^(-1)(T1 U T2) if and only if x belongs to f^(-1)(T1) U f^(-1)(T2), we can conclude that f^(-1)(T1 U T2) = f^(-1)(T1) U f^(-1)(T2).
b. Now, using a similar approach, we will prove that f^(-1)(T1 ∩ T2) = f^(-1)(T1) ∩ f^(-1)(T2).
Proof:
Let x be an element in f^(-1)(T1 ∩ T2). By definition, this means that f(x) belongs to T1 ∩ T2.
This implies that f(x) belongs to T1 and f(x) belongs to T2.
By definition, x belongs to f^(-1)(T1) and x belongs to f^(-1)(T2). Therefore, x belongs to f^(-1)(T1) ∩ f^(-1)(T2).
Now, let x be an element in f^(-1)(T1) ∩ f^(-1)(T2). This means that x belongs to both f^(-1)(T1) and f^(-1)(T2).
By definition of inverse image, f(x) belongs to T1 and f(x) belongs to T2, which implies that f(x) belongs to T1 ∩ T2.
Hence, by both inclusions, we can conclude that f^(-1)(T1 ∩ T2) = f^(-1)(T1) ∩ f^(-1)(T2).
c. Next, we will prove that f^(-1)(T1) - f^(-1)(T2) = f^(-1)(T1 - T2).
Proof:
To show that two sets are equal, we need to prove that each side is contained in the other.
Part 1: Let x be an element in f^(-1)(T1) - f^(-1)(T2). This means that x belongs to f^(-1)(T1), but x does not belong to f^(-1)(T2).
By definition, this implies that f(x) belongs to T1, but f(x) does not belong to T2.
Therefore, f(x) belongs to T1 - T2, which implies x belongs to f^(-1)(T1 - T2).
Part 2: Now, let x be an element in f^(-1)(T1 - T2). By definition, this means that f(x) belongs to T1 - T2.
Hence, f(x) belongs to T1, but it does not belong to T2. This implies that x belongs to f^(-1)(T1), but x does not belong to f^(-1)(T2).
Therefore, both inclusions are satisfied, and we can conclude that f^(-1)(T1) - f^(-1)(T2) = f^(-1)(T1 - T2).
d. Lastly, we will prove that if T1 is a proper subset of T2, then f^(-1)(T1) is a proper subset of f^(-1)(T2).
Proof:
Let T1 be a proper subset of T2. This implies that there exists an element y in T2 which does not belong to T1.
Now, let x be an element in f^(-1)(T1). By definition, f(x) belongs to T1.
Since T1 is a proper subset of T2, f(x) does not belong to T2. Hence, x does not belong to f^(-1)(T2).
Therefore, every element in f^(-1)(T1) is not in f^(-1)(T2), so f^(-1)(T1) is a proper subset of f^(-1)(T2) when T1 is a proper subset of T2.
I hope this helps you understand the concepts of inverse mappings and how to prove the given statements!