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Chapter 1: Problem 13

Prove: An isolated point of \(S\) is a boundary point of \(S^{c}\).

Short Answer

Expert verified
Question: Prove that an isolated point of a set \(S\) is a boundary point of its complement \(S^c\). Answer: To prove that an isolated point of \(S\) is a boundary point of its complement \(S^c\), we need to show that the isolated point is in the closure of both \(S\) and \(S^c\). By definition, an isolated point of a set has a neighborhood with no other points of the set in it. Therefore, any neighborhood of the isolated point will intersect both the set \(S\) and the complement \(S^c\), making the isolated point a boundary point of the complement \(S^c\).

Step by step solution

01

Definitions

An isolated point of a set \(S\) is a point that is in \(S\) but not a limit point of \(S\). A boundary point of a set \(S\) is a point that belongs to the closure of both \(S\) and its complement \(S^c\). The complement of a set \(S\), denoted as \(S^c\), is the set of all points not in \(S\).
02

Proving an isolated point of \(S\) is in the closure of \(S^{c}\)

Let \(x\) be an isolated point of \(S\). It means there exists a neighborhood \(N(x)\) such that \(N(x) \cap S = \{x\}\). Since \(x\) is not a limit point of \(S\), there exists a positive distance \(r\) between \(x\) and other points in \(S\). Therefore, for every neighborhood \(N(x)\) of radius \(r\), we have \(N(x) \cap S^c \neq \emptyset\). Thus, \(x\) is in the closure of \(S^c\).
03

Proving an isolated point of \(S\) is in the closure of \(S\)

Since \(x\) is an isolated point of \(S\), there exists a neighborhood \(N(x)\) such that \(N(x) \cap S = \{x\}\). This means that \(x\) is in the closure of \(S\).
04

Conclusion

Since an isolated point \(x\) is in the closure of both \(S\) and \(S^c\), it is a boundary point of \(S^c\). This completes the proof.

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Most popular questions from this chapter

Find the supremum and infimum of each \(S .\) State whether they are in \(S\). (a) \(S=\left\\{x \mid x=-(1 / n)+\left[1+(-1)^{n}\right] n^{2}, n \geq 1\right\\}\) (b) \(S=\left\\{x \mid x^{2}<9\right\\}\) (c) \(S=\left\\{x \mid x^{2} \leq 7\right\\}\) (d) \(S=\\{x|| 2 x+1 \mid<5\\}\) (e) \(S=\left\\{x \mid\left(x^{2}+1\right)^{-1}>\frac{1}{2}\right\\}\) (f) \(S=\left\\{x \mid x=\right.\) rational and \(\left.x^{2} \leq 7\right\\}\)

Suppose that \(a_{1} \leq a_{2} \leq \cdots \leq a_{n}\) and \(b_{1} \leq b_{2} \leq \cdots \leq b_{n} .\) Let \(\left\\{\ell_{1}, \ell_{2}, \ldots \ell_{n}\right\\}\) be a permutation of \(\\{1,2, \ldots, n\\},\) and define $$ Q\left(\ell_{1}, \ell_{2}, \ldots, \ell_{n}\right)=\sum_{i=1}^{n}\left(a_{i}-b_{\ell_{i}}\right)^{2} $$ Show that $$ Q\left(\ell_{1}, \ell_{2}, \ldots, \ell_{n}\right) \geq Q(1,2, \ldots, n) $$

Let \(P_{n}\) be the proposition that $$ 1+2+\cdots+n=\frac{(n+2)(n-1)}{2} $$ (a) Show that \(P_{n}\) implies \(P_{n+1}\). (b) Is there an integer \(n\) for which \(P_{n}\) is true?

Suppose that \(m\) and \(n\) are integers, with \(0 \leq m \leq n .\) The binomial coefficient \(\left(\begin{array}{l}n \\ m\end{array}\right)\) is the coefficient of \(t^{m}\) in the expansion of \((1+t)^{n} ;\) that is, $$ (1+t)^{n}=\sum_{m=0}^{n}\left(\begin{array}{l} n \\ m \end{array}\right) t^{m} $$ From this definition it follows immediately that $$ \left(\begin{array}{l} n \\ 0 \end{array}\right)=\left(\begin{array}{l} n \\ n \end{array}\right)=1, \quad n \geq 0 $$ For convenience we define $$ \left(\begin{array}{r} n \\ -1 \end{array}\right)=\left(\begin{array}{c} n \\ n+1 \end{array}\right)=0, \quad n \geq 0 $$

Find and prove by induction an explicit formula for \(a_{n}\) if \(a_{1}=1\) and, for \(n \geq 1,\) (a) \(a_{n+1}=\frac{a_{n}}{(n+1)(2 n+1)}\) (b) \(a_{n+1}=\frac{3 a_{n}}{(2 n+2)(2 n+3)}\) (c) \(a_{n+1}=\frac{2 n+1}{n+1} a_{n}\) (d) \(a_{n+1}=\left(1+\frac{1}{n}\right)^{n} a_{n}\)
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