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Chapter 2: Problem 12

Show that \(f: A \rightarrow B\) is continuous if and only if for every closed set \(F\) relative to \(B, f^{-1}(F)\) is a closed set relative to \(A\).

Short Answer

Expert verified
For the forward implication, we showed that the pre-image of the complement of a closed set under a continuous function is open, which implies that the pre-image of the closed set is closed. For the backward implication, we showed that if the pre-image of the complement of an open set (which is a closed set) is closed, then the pre-image of the open set is open, which implies that the function is continuous. Therefore, a function is continuous if and only if the pre-image of every closed set is closed.

Step by step solution

01

Prove the forward implication

Assume \(f: A \rightarrow B\) is continuous. Let \(F\) be a closed set in \(B\). We need to show that \(f^{-1}(F)\) is closed in \(A\). The complement of \(F\) in \(B\), denoted \(B \setminus F\), is an open set since \(F\) is closed. By definition of continuous functions, the pre-image of an open set under a continuous function is open. Therefore, \(f^{-1}(B \setminus F)\) is open in \(A\). This is the complement of \(f^{-1}(F)\) in \(A\), so by definition of closed sets, \(f^{-1}(F)\) is closed in \(A\). Thus, the forward implication is proven.
02

Prove the backward implication

Assume that for every closed set \(F\) relative to \(B, f^{-1}(F)\) is a closed set relative to \(A\). We want to show that \(f\) is continuous. Let \(G\) be an open set in \(B\). Note that \(B \setminus G\) is a closed set in \(B\). By assumption, \(f^{-1}(B \setminus G)\) is a closed set in \(A\). This means the complement of \(f^{-1}(B \setminus G)\) in \(A\), is open. But the complement of \(f^{-1}(B \setminus G)\) in \(A\) is \(f^{-1}(G)\), so we have that the pre-image of an open set under \(f\) is open, which by definition means that \(f\) is continuous. Thus, the backward implication is proven.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Closed Sets
In topology, a closed set is best understood as a set that contains all its limit points.
This means any boundary point of the set belongs to the set itself.
Closed sets have a complementary relationship with open sets.
  • For example, if a set is closed in a certain topological space, its complement (i.e., everything outside of this set) will be open in that same space.
  • This relationship is crucial because it allows for the use of complement properties in proofs and discussions regarding topological structures.
In the context of a function between two topological spaces, the concept of closed sets helps us analyze the behavior of a function even when direct handling is complicated.
By understanding closed sets, we can make inferences about the function's continuity by looking at the pre-images of these closed sets in the function's domain.
Pre-image
The concept of a pre-image arises when analyzing functions between two sets, such as two topological spaces.
The pre-image of a set under a function is essentially the set of all elements in the domain that map to the set in the range through the function.
It is represented by the notation \( f^{-1} \).
  • For example, if we have a function \( f: A \rightarrow B \), and a set \( F \) in the space \( B \), then the pre-image \( f^{-1}(F) \) is the set of all elements in \( A \) that \( f \) maps into \( F \).
  • The pre-image concept is especially useful for studying the properties of functions, such as continuity.
In our exercise, understanding pre-images allows us to determine the behavior of closed sets when a function is applied
This is because the nature of pre-images, in relation to set operations like union and intersection, preserve certain properties across the mapping.
Continuous Functions
A continuous function is a fundamental concept in topology and analysis, capturing the intuitive idea that small changes in input result in small changes in output.
In topological terms, a function is continuous if the pre-image of every open set in the range is open in the domain.
  • Alternatively, this can be characterized using closed sets: if the pre-image of every closed set in the range is closed in the domain.
  • This alignment with open and closed sets is what allows topologists to understand the function's behavior at a deeper level.
From our exercise, the continuity of a function \( f: A \rightarrow B \) is verified through the behavior of pre-images:
  • If the pre-image of every closed set in \( B \) is also closed in \( A \, \) then the function is continuous.
  • This mirrors the idea that continuity allows for the preservation of structure under the function's mapping.
Thus, understanding these conditions helps in proving or disproving whether a given function adheres to the continuity requirement.

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Most popular questions from this chapter

Show that a compact set of real numbers contains its greatest lower bound and its least upper bound. Can this occur for a set of real numbers that is not compact?

If \(\left\\{S_{n}\right\\}\) is a sequence of nonempty compact sets with \(S_{n+1} \subset S_{n}\) for every \(n\), show that \(\bigcap_{n=1}^{\infty} S_{n} \neq \phi\).

Prove that \(f(z)=1 /(1-z)\) is not uniformly continuous for \(|z|<1\).

Discuss continuity and uniform continuity for the following functions. (a) \(f(z)=\frac{1}{1-z} \quad(|z|<1)\) (b) \(f(z)=\frac{1}{z} \quad(|z| \geq 1)\) (c) \(f(z)=\left\\{\begin{array}{ll}\frac{|z|}{z} & \text { if } 0<|z| \leq 1 \\\ 0 & \text { if } z=0\end{array}\right.\) (d) \(f(z)=\left\\{\begin{array}{ll}\frac{\operatorname{Re} z}{z} & \text { if } 0<|z|<1 \\ 1 & \text { if } z=0 .\end{array}\right.\)

Show that \(\left\\{z_{n}\right\\}\) approaches \(\infty\) if and only if \(\left\\{\left|z_{n}\right|\right\\}\) approaches \(\infty\).
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