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Chapter 11: Problem 2

Show that the intervals \((a, \infty)\) and \((-\infty, a)\) are open sets, and that the intervals \([b, \infty)\) and \((-\infty, b]\) are closed sets.

Short Answer

Expert verified
The intervals (a, 鈭�) and (-鈭�, a) are open sets, and [b, 鈭�) and (-鈭�, b] are closed sets.

Step by step solution

01

Check if the intervals (a, 鈭�) and (-鈭�, a) are open sets

Select an arbitrary point \(x\) in the interval \((a, 鈭�)\). Now we will find an epsilon (\(蔚\)) such that \((x - 蔚, x + 蔚)\) is contained within \((a, 鈭�)\). Let鈥檚 set \(蔚 = x - a\) which means \(蔚 > 0\) as \(x > a\). This shows that for any \(x\) there exists a \(蔚 > 0\) such that \((x - 蔚, x + 蔚)\) is contained within \((a, 鈭�)\), hence \((a, 鈭�)\) is an open set. Use a similar approach for the interval \((-鈭�, a)\). Any point \(x\) in \((-鈭�, a)\) will have an epsilon-neighbourhood entirely fit within the set. Hence, \((-鈭�, a)\) is an open set.
02

Check if the intervals [b, 鈭�) and (-鈭�, b] are closed sets

We now show that [b, 鈭�) is a closed set by showing that its complement is an open set. The complement of [b, 鈭�) is (-鈭�, b), which is already shown as an open set. Hence [b, 鈭�) is a closed set. Similarly, the complement of the interval (-鈭�, b] is (b, 鈭�), which is an open set. Therefore, (-鈭�, b] is a closed set.

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Most popular questions from this chapter

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

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.

If \((S, d)\) is a metric space, a subset \(A \subseteq S\) is said to be bounded if there exists \(x_{0} \in S\) and a number \(B>0\) such that \(A \subseteq\left\\{x \in S: d\left(x, x_{0}\right) \leq B\right\\}\). Show that if \(A\) is a compact subset of \(S\), then \(A\) is closed and bounded.

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.

Show that the set \(\mathbb{Q}\) of rational numbers is neither open nor closed.
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