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Chapter 11: Problem 11

Show that a set \(G \subseteq \mathbb{R}\) is open if and only if it does not contain any of its boundary points.

Short Answer

Expert verified
Given the definitions of open set and boundary points, it has been proved that if a set \(G\) does not contain any of its boundary points, it is open; and if \(G\) is open, it does not contain any of its boundary points.

Step by step solution

01

Recall the Definition of an Open Set and Boundary Point

An open set \(G\) in \(\mathbb{R}\) is a set which, for every \(x \in G\), there exists some \(\epsilon > 0\) so that \(B(x, \epsilon) = {y \in \mathbb{R}: |x-y|<\epsilon}\) is completely contained within \(G\). A boundary point of \(G\), is a point that every open set containing it intersects both \(G\) and its complement \(\mathbb{R} \setminus G\).
02

Prove that an Open Set Does Not Contain its Boundary Points

Assuming \(G\) is an open set and \(x\) is a boundary point. It means that every open set containing \(x\) intersects both \(G\) and \(\mathbb{R} \setminus G\). Given \(G\) is an open set, there exists an \(\epsilon > 0\) such that \(B(x, \epsilon)\) is completely in \(G\). But this contradicts the assumption that \(x\) is a boundary point, as \(B(x, \epsilon)\) does not intersect \(\mathbb{R} \setminus G\). Therefore, \(G\) cannot contain its boundary points if it is open.
03

Prove that if a Set Does Not Contain its Boundary Points, it is Open

Assume that \(G\) does not contain its boundary points and choose an arbitrary \(x \in G\). By assumption, \(x\) cannot be a boundary point. So, there is some \(\epsilon > 0\) such that either \(B(x, \epsilon)\) is completely inside \(G\), or \(B(x, \epsilon)\) is completely inside \(\mathbb{R} \setminus G\). As \(x \in G\), \(B(x, \epsilon)\) cannot be in \(\mathbb{R} \setminus G\), thus \(B(x, \epsilon)\) must be completely in \(G\). This satisfies the definition of an open set.

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Most popular questions from this chapter

Let \(K \neq \emptyset\) be a compact set in \(\mathbb{R}\). Show that inf \(K\) and sup \(K\) exist and belong to \(K\).

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.

Let \(I:=[1, \infty)\) and let \(f(x):=\sqrt{x-1}\) for \(x \in I\). For each \(\varepsilon\) -neighborhood \(G=(-\varepsilon .+\varepsilon)\) of 0\. exhibit an open set \(H\) such that \(H \cap I=f^{-1}(G)\).

Show that the set \(\mathbb{N}\) of natural numbers is a closed set.

Show that if \(G\) is an open set and \(F\) is a closed set, then \(G \backslash F\) is an open set and \(F \backslash G\) is a closed set.
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