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Chapter 11: Problem 12
Show that a set \(F \subseteq \mathbb{R}\) is closed if and only if it contains all of its boundary points.
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Prove that in any metric space, an \(\varepsilon\) -neighborhood of a point is an open set.
Show that if \(f: \mathbb{R} \rightarrow \mathbb{R}\) is continuous. then the set \(\\{x \in \mathbb{R}: f(x) \leq \alpha\\}\) is closed in \(\mathbb{R}\) for each \(\alpha \in \mathbb{R}\)
Give an example of a function \(f: \mathbb{R} \rightarrow \mathbb{R}\) such that the set \(\\{x \in \mathbb{R}: f(x)=1\\}\) is neither open nor closed in \(\mathbb{R}\).
A point \(x \in \mathbb{R}\) is said to be a boundary point of \(A \subseteq \mathbb{R}\) in case every neighborhood \(V\) of \(x\) contains points in \(A\) and points in \(\mathcal{C}(A)\). Show that a set \(A\) and its complement \(\mathcal{C}(A)\) have exactly the same boundary points.
Prove, using Definition \(11.2 .2\), that if \(K_{1}\) and \(K_{2}\) are compact sets in \(\mathbb{R}\), then their union \(K_{1} \cup K_{2}\) is compact.
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