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Chapter 11: Problem 9

Prove that in any metric space, an \(\varepsilon\) -neighborhood of a point is an open set.

Short Answer

Expert verified
In any metric space, an \(\varepsilon\) -neighborhood of a point is indeed an open set. This is proven by showing that for every point in the \(\varepsilon\) -neighborhood there exists another \(\varepsilon_1\) -neighborhood which is completely contained in the \(\varepsilon\) -neighborhood. Hence, every point in the \(\varepsilon\) -neighborhood is an interior point which makes the \(\varepsilon\) -neighborhood an open set.

Step by step solution

01

Define \(\varepsilon\) -neighborhood and open set

An \(\varepsilon\) -neighborhood of a point \(x\) in a metric space \(X\) is the set of all points \(y\) in \(X\) such that the distance between \(x\) and \(y\) is less than \(\varepsilon\), where \(\varepsilon > 0\). An open set in a metric space is a set where every point is an interior point.
02

Pick a point in the \(\varepsilon\) -neighborhood

Let's pick any point \(y\) in the \(\varepsilon\) -neighborhood of \(x\). By the definition of \(\varepsilon\) -neighborhood, the distance between \(x\) and \(y\) is less than \(\varepsilon\). Denote this distance as \(d\).
03

Define a new \(\varepsilon_1\)

Let \(\varepsilon_1 = \varepsilon - d > 0\). \(\varepsilon_1\) is the distance from \(y\) to the boundary of the \(\varepsilon\) -neighborhood of \(x\).
04

Show that every point in the \(\varepsilon_1\) -neighborhood of \(y\) is in the \(\varepsilon\) -neighborhood of \(x\)

Let's pick an arbitrary point \(z\) in the \(\varepsilon_1\) -neighborhood of \(y\). The distance from \(y\) to \(z\) is less than \(\varepsilon_1\). By the triangle inequality, the distance from \(x\) to \(z\) is less than or equal to the distance from \(x\) to \(y\) plus the distance from \(y\) to \(z\), which is less than \(d + \varepsilon_1 = \varepsilon\). Therefore, \(z\) is in the \(\varepsilon\) -neighborhood of \(x\).
05

Prove that every point in the \(\varepsilon\) -neighborhood is an interior point

By the previous step, we've shown that for every point \(y\) in the \(\varepsilon\) -neighborhood of \(x\), there exists an \(\varepsilon_1\) -neighborhood of \(y\) which is a subset of the \(\varepsilon\) -neighborhood of \(x\). Therefore, every point in the \(\varepsilon\) -neighborhood of \(x\) is an interior point.
06

Conclusion

Since every point in the \(\varepsilon\) -neighborhood of \(x\) is an interior point, the \(\varepsilon\) -neighborhood of \(x\) is an open set. This proves our claim that in any metric space, an \(\varepsilon\) -neighborhood of a point is an open set.

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Most popular questions from this chapter

Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be defined by \(f(x)=x^{2}\) for \(x \in \mathbb{R}\) (a) Show that the inverse image \(f^{-1}(I)\) of an open interval \(I:=(a, b)\) is either an open interval. the union of two open intervals, or empty, depending on \(a\) and \(b\). (b) Show that if \(I\) is an open interval containing 0 , then the direct image \(f(I)\) is not open.

If \(A \subseteq \mathbb{R}\), let \(A^{\circ}\) be the union of all open sets that are contained in \(A ;\) the set \(A^{\circ}\) is called the interior of \(A\). Show that \(A^{\circ}\) is an open set, that it is the largest open set contained in \(A\), and that a point \(z\) belongs to \(A^{\circ}\) if and only if \(z\) is an interior point of \(A\).

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.

Show that \(A=\\{1 / n: n \in \mathbb{N}\\}\) is not a closed set, but that \(A \cup\\{0\\}\) is a closed set.

Exhibit an open cover of \(\mathbb{N}\) that has no finite subcover.
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