Question

Suppose that \(f:(A, \rho) \rightarrow(B, \sigma)\) and \(D_{f}=A\). Show that the following statements are equivalent. (a) \(f\) is continuous on \(A\). (b) If \(V\) is any open set in \((B, \sigma)\), then \(f^{-1}(V)\) is open in \((A, \rho)\). (c) If \(V\) is any closed set in \((B, \sigma)\), then \(f^{-1}(V)\) is closed in \((A, \rho)\).

Short answer

Question: Prove that the following statements about the continuity of a function \(f\) from a topological space \((A,\rho)\) to a topological space \((B,\sigma)\) are equivalent: (a) \(f\) is continuous on \(A\) (b) \(f^{-1}(V)\) is open in \((A, \rho)\) for any open set \(V\) in \((B, \sigma)\) (c) \(f^{-1}(V)\) is closed in \((A, \rho)\) for any closed set \(V\) in \((B, \sigma)\) Answer: The three statements (a), (b), and (c) are equivalent because: 1. If \(f\) is continuous on \(A\), then \(f^{-1}(V)\) is open in \((A, \rho)\) for any open set \(V\) in \((B, \sigma)\). 2. If \(f^{-1}(V)\) is open in \((A, \rho)\) for any open set \(V\) in \((B, \sigma)\), then \(f^{-1}(V)\) is closed in \((A, \rho)\) for any closed set \(V\) in \((B, \sigma)\). 3. If \(f^{-1}(V)\) is closed in \((A, \rho)\) for any closed set \(V\) in \((B, \sigma)\), then \(f\) is continuous on \(A\). The proofs of these implications can be found in the provided solution.

Step-by-step solution

Proof: (a) implies (b)

Let \(f\) be continuous on \(A\). To show that \(f^{-1}(V)\) is open in \((A, \rho)\) for any open set \(V\) in \((B, \sigma)\), let \(V\) be an open set in \((B, \sigma)\) and consider \(f^{-1}(V)\). Since \(f\) is continuous, for any point \(x \in f^{-1}(V)\), there exists an open neighborhood \(U_x\) of \(x\) in \((A, \rho)\) such that \(f(U_x) \subseteq V\). Thus, we have \(U_x \subseteq f^{-1}(V)\). Therefore, we can write \(f^{-1}(V) = \bigcup_{x \in f^{-1}(V)} U_x\), which is an open set in \((A, \rho)\) since it is the union of open sets.

Proof: (b) implies (c)

Let \(f^{-1}(V)\) be open in \((A, \rho)\) for any open set \(V\) in \((B, \sigma)\). Now, let \(V\) be any closed set in \((B, \sigma)\). Then, the complement of \(V\), denoted \(V^c\), is an open set in \((B, \sigma)\). So, \(f^{-1}(V^c)\) is an open set in \((A, \rho)\). The complement of \(f^{-1}(V^c)\) in \((A, \rho)\), denoted as \((f^{-1}(V^c))^c\), is equal to \(f^{-1}(V)\) (by the properties of the inverse image). Since the complement of an open set is closed, \(f^{-1}(V)\) must be closed in \((A, \rho)\).

Proof: (c) implies (a)

Finally, let \(f^{-1}(V)\) be closed in \((A, \rho)\) for any closed set \(V\) in \((B, \sigma)\). To show that \(f\) is continuous on \(A\), it is sufficient to prove that for any point \(x \in A\), and any open neighborhood \(V\) of \(f(x)\) in \((B, \sigma)\), there exists an open neighborhood \(U\) of \(x\) in \((A, \rho)\) such that \(f(U) \subseteq V\). Let \(W = V^c\), which is a closed set in \((B, \sigma)\). Then \(f^{-1}(W)\) is closed in \((A, \rho)\). Since \(x \notin f^{-1}(W)\), \(A \setminus f^{-1}(W)\) is an open set containing \(x\). Let \(U\) be an open subset of \(A \setminus f^{-1}(W)\) that contains \(x\). Then, \(f(U) \subseteq V\), proving that \(f\) is continuous on \(A\). Now all implications have been proven, which proves that the statements (a), (b), and (c) are equivalent.

Key concepts

Inverse Image

When we talk about the inverse image (or preimage) in the context of a function, we're discussing how we can trace back from the destination set to the starting set. Specifically, for any given set in the codomain, the inverse image consists of all elements from the domain that are mapped to that set. Let's break it down further:

• Imagine you have a function \( f: A \to B \). If \( V \) is a subset of \( B \), the inverse image of \( V \), denoted \( f^{-1}(V) \), is the collection of all elements in \( A \) that \( f \) maps into \( V \).
• This is useful in understanding function behavior since it tells us where in the domain our interest set in the codomain comes from.
• Importantly, the properties of inverse images, like open and closed characteristics, are crucial in proving continuity and other functional properties.

Understanding inverse images is foundational when analyzing how functions interact with topological properties like "open" and "closed" sets.

Open Sets

Open sets are a mainstay of topology, deeply tied to the concept of continuity in functions. In simple terms, an open set is a set that doesn't include its boundary. When we explore continuous functions, we can understand this concept better through the role of open sets:

• If a function \( f: A \to B \) is continuous, and \( V \) is an open set in \( B \), then the inverse image \( f^{-1}(V) \) should be open in \( A \). This property is a hallmark of continuity.
• Why does this matter? Because it ensures that the function behaves "smoothly," without any abrupt jumps or interruptions as you move from one point to another within the domain.
• This property is what helps us rigorously define continuity beyond elementary or intuitive notions.

So when you're dealing with continuous functions, keep in mind how open sets maintain their open nature through inverse images.

Closed Sets

Closed sets complement open sets, both in mathematical theory and in practice. A closed set includes its boundary, and this characteristic is just as valuable in discussions around functions and continuity.

• For a function \( f: A \to B \), if every closed set \( V \) in \( B \) results in a closed inverse image \( f^{-1}(V) \) in \( A \), the function is continuous.
• This property provides a useful way to determine continuity by checking the inverse images of closed sets, as a function's ability to "properly wrap" closed sets is a sign of well-behaved movement between domain and codomain.
• While open sets are directly linked with continuity by their definition, closed sets offer an alternative perspective. By examining them, we often find simpler or more accessible proofs in various scenarios.

Overall, closed sets offer a reliable method for studying continuity, particularly when paired with the idea of complementarity between open and closed sets.