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Chapter 11: Problem 7

Show that if \(d\) is the discrete metric on a set \(S\), then every subset of \(S\) is both open and closed in \((S, d)\).

Short Answer

Expert verified
The proof is centered around the unique property of the discrete space, where the distance between any two points is either 0 or 1. This allows us to construct an open ball for any point in a subset of the metric space that only contains that point. As a result, all subsets of the metric space are shown to be both open and closed.

Step by step solution

01

Understand open sets in metric spaces

A set \(A\) in a metric space \(X\) is said to be open if, for every point \(x\) in \(A\), there is a radius \(r > 0\) such that the open ball \(B_{r}(x)\) is contained within \(A\). In other words, for every point in the set, there is always 'room' to move in the set, without leaving it.
02

Show that every subset is open in the discrete metric space

Consider an arbitrary subset \(A\) of \(S\) and any point \(x\) in \(A\). We have to show that there exists a radius \(r > 0\) such that the open ball \(B_{r}(x)\) is contained within \(A\). As we know, the open ball \(B_{r}(x)\) is defined to be the set \{ \(y \in S | d(y, x) < r \}\). For the discrete metric, since \(d(y, x)\) can only be 0 or 1, we can take any \(r\) between 0 and 1, say, 0.5. Then, \(B_{0.5}(x)\) only includes the point \(x\) itself, which is definitely in \(A\). Therefore, \(A\) is open.
03

Understand closed sets in metric spaces

A set \(B\) in a metric space \(X\) is said to be closed if its complement \(X\B\) is open. In other words, if all points that are not in \(B\) form an open set, then \(B\) is considered to be closed.
04

Show that every subset is closed in the discrete metric space

Now, consider again an arbitrary subset \(A\) of \(S\) and consider its complement \(A^{c} = S\A\). Following the reasoning of Step 2, we can again take any \(r\) between 0 and 1, say, 0.5. For any \(x\) in \(A^{c}\), the open ball \(B_{0.5}(x)\) will only include \(x\) itself and hence is a subset of \(A^{c}\). Therefore, \(A^{c}\) is open, and consequently \(A\) is closed. This shows that every subset of \(S\) is both open and closed.

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Most popular questions from this chapter

If \(K_{1}\) and \(K_{2}\) are disjoint nonempty compact sets, show that there exist \(k_{i} \in K_{i}\) such that \(0<\left|k_{1}-k_{2}\right|=\inf \left\\{\left|x_{1}-x_{2}\right|: x_{1} \in K_{i}\right\\}\)

If \(A \subseteq \mathbb{R}\), let \(A^{-}\) be the intersection of all closed sets containing \(A\); the set \(A^{-}\) is called the closure of \(A\). Show that \(A^{-}\) is a closed set, that it is the smallest closed set containing \(A\), and that a point \(w\) belongs to \(A^{-}\) if and ondy if \(w\) is either an interior point or a boundary point of \(A\).

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

Use the Heine-Borel Theorem to prove the following version of the Bolzano- Weierstrass Theorem: Every bounded infinite subset of \(\mathbb{R}\) has a cluster point in \(\mathbb{R}\). (Note that if a sct has no cluster points, then it is closed by Theorem \(11.1 .8 .\) )

Using the notation of the preceding exercise, let \(A, B\) be sets in \(\mathbb{R}\). Show that we have \(A \subseteq\) \(A^{-},\left(A^{-}\right)^{-}=A^{-}\), and that \((A \cup B)^{-}=A^{-} \cup B^{-}\). Show that \((A \cap B)^{-} \subseteq A^{-} \cap B^{-}\), and give an example to show that the inclusion may be proper.
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