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Chapter 11: Problem 8

Prove that the intersection of an arbitrary collection of compact sets in \(\mathbb{R}\) is compact.

Short Answer

Expert verified
The intersection of an arbitrary collection of compact sets in \(\mathbb{R}\) is also compact because it is both closed and bounded, satisfying the characteristics of a compact set.

Step by step solution

01

Checking Closed Property

Consider an arbitrary collection of compact sets \(C_1, C_2, C_3, ..., C_n\) in \(\mathbb{R}\). The intersection of these compact sets can be represented as \(\cap_{i=1}^{n} C_i\). Since each of these compact sets \(C_i\) is closed, the intersection of these, which is also a set, is therefore closed, because the intersection of any collection of closed sets is always closed.
02

Prove Boundedness

The compact sets are also bounded, which means there exist two real numbers \(a\) and \(b\) such that every number within the set lies between \(a\) and \(b\). Given that each of the compact sets \(C_i\) are bounded, there must be a maximum upper bound and minimum lower bound among all the sets. These bounds will also serve as the bounds for \(\cap_{i=1}^{n} C_i\) and prove that the intersection is bounded.
03

Conclusion

To conclude, since the intersection \(\cap_{i=1}^{n} C_i\) is both closed and bounded, according to Heine-Borel Theorem, it is a compact set as well. Therefore, we have shown that the intersection of an arbitrary collection of compact sets in \(\mathbb{R}\) is also compact.

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Most popular questions from this chapter

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 \(A\).

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 \(\alpha \in \mathbb{R}\)

Let \(h: \mathbb{R} \rightarrow \mathbb{R}\) be defined by \(h(x):=1\) if \(0 \leq x \leq 1, h(x):=0\) otherwise. Find an open set \(G\) such that \(h^{-1}(G)\) is not open, and a closed set \(F\) such that \(h^{-1}(F)\) is not closed.

Prove, using Definition 11.2.2. that if \(F\) is a closed subset of a compact set \(K\) in \(\mathbb{R}\), then \(F\) is compact.

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 \(11.1 .8 .\) )
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