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Chapter 11: Problem 9

Prove that \(f: \mathbb{R} \rightarrow \mathbb{R}\) is continuous if and only if for each closed set \(F\) in \(\mathbb{R}\), the inverse image \(f^{-1}(F)\) is closed.

Short Answer

Expert verified
The proof consists of two parts. First, assuming \(f\) is continuous, it is shown that for any closed set \(F\), \(f^{-1}(F)\) is closed. Second, assuming the pre-image of any closed set is closed, it is shown that \(f\) is continuous.

Step by step solution

01

Proving that continuity implies pre-images of closed sets are closed

Assume \(f\) is continuous. Let \(F\) be a closed set. The definition of continuity states that for \( \forall \epsilon > 0, \exists \delta > 0 \) such that if |x - a| < \( \delta \), then |f(x) - f(a)| < \( \epsilon \). If \(F\) is closed, then its complement, \( \mathbb{R} \setminus F \), is open. As the image of an open set under a continuous function is open, \(f^{-1}( \mathbb{R} \setminus F )\) is open. But then by definition, its complement \(f^{-1}(F)\) must be closed.
02

Proving that if pre-images of closed sets are closed the function is continuous

If for every closed set \(F \subset \mathbb{R}\), \(f^{-1}(F)\) is closed, we want to show that \(f\) is continuous. We use the fact that a function is continuous if and only if, for every open set \(V \subseteq \mathbb{R}\), the preimage \(f^{-1}(V)\) is open. Given any open set \(V\), its complement \(\mathbb{R} \setminus V\) is closed. Thus by hypothesis, \(f^{-1}( \mathbb{R} \setminus V )\) is closed. Hence its complement \(f^{-1}(V)\), being the complement of a closed set, is open. Therefore, \(f\) is continuous.

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