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Chapter 8: Problem 10

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 set S^c. Answer: To prove that an isolated point of a set S is a boundary point of its complement set S^c, we need to show that every neighborhood of the isolated point contains points from both S and S^c. By definition, an isolated point of the set S has a neighborhood containing only the point itself and no other points from set S. Therefore, any neighborhood of the isolated point contains points from S^c (points not in S), as well as the isolated point from S. This satisfies the definition of a boundary point, proving that an isolated point of a set S is indeed a boundary point of its complement set S^c.

Step by step solution

01

Definitions

Recall the following definitions. - Isolated Point: A point x in a set S is called an isolated point if there exists a neighborhood of x such that the neighborhood contains no points of the set S other than x. - Boundary Point: A point x is called a boundary point of a set S if it is in any neighborhood of x, the neighborhood has points from both the set S and its complement, S^c. Let's prove that an isolated point of the set S is a boundary point of its complement set S^c.
02

Assume x is an isolated point of S

Let x be an isolated point of the set S. By definition, there exists a neighborhood around x, containing x but no other points from set S.
03

Show that x is a boundary point of S^c

To prove x is a boundary point of S^c, we have to show that every neighborhood of x contains points from both S^c and S. Given that x is an isolated point of S, we know that there exists a neighborhood, N, containing x but no other points from S. Now consider an arbitrary neighborhood, M, of x. Since x is in M, the intersection of M with S is nonempty (as it contains at least the point x itself). Now, let's see whether each neighborhood M of x contains points from S^c. Since x is part of the set S, then we know that x is not a part of the set S^c, hence x itself cannot be considered as a shared point between S^c and M. However, we know that the intersection N鈭㏒ contains only x, meaning there are no other points from the set S in N. Since any neighborhood M of x must overlap with N at least partially (as they both contain x), it means any neighborhood M also contains elements from N that do not belong to the set S, as there is always a portion shared between M and N that doesn't include x. Hence, the neighborhood M contains non-intersecting points from both S (x) and S^c (the remaining portion of the overlapping N). Thus, by definition, x is a boundary point of the set S^c. We have now shown that an isolated point of a set S, in this case x, is a boundary point of S^c.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Isolated Point
Imagine you're standing alone in a vast field with no one around for miles鈥攖hat's similar to being an isolated point in the realm of real analysis. An isolated point of a set, say set S, is like you in that field: it has some space around it where no other members of the set lie. More formally, if we pick a point x from set S, it鈥檚 isolated when there is a neighborhood around x, a sort of bubble, within which there are no other points of S except for x itself.

Here鈥檚 why this is important: isolated points help us understand the overall 'shape' and characteristics of sets in mathematics. They're like the individual pieces in a puzzle that do not fit with the rest, offering unique insights into the nature of the set they belong to. However, do not mix them up with boundary points, as they are an entirely different concept which we will explore next.
Complement Set
The idea of a complement set is straightforward鈥攖hink of it as the 'opposite day' for sets. In the realm of real analysis, the complement is essentially everything that is not included in your set of interest. For a given set S, the complement set, often denoted by Sc, contains all the elements that are not part of S.

To visualize this, consider a collection of red balls (set S) in a box full of various colored balls. If you take out all the red balls, the collection of all the non-red balls left in the box will represent your complement set, Sc. Complement sets are a fundamental concept in set theory and are used in proving various mathematical properties. For instance, understanding how an isolated point in a set S relates to the boundary points of the complement set Sc sheds light on the interconnectedness within mathematical structures.
Neighborhood in Real Analysis
In your neighborhood, there's your home, maybe a park down the street, and a grocery store around the corner. In real analysis, a neighborhood is somewhat similar; it's the mathematical equivalent of a local area around a point. More precisely, for any point x, a neighborhood is a set of points that lie within a certain distance from x. This distance is often represented by a positive value called the radius of the neighborhood.

Why is this concept useful? Well, neighborhoods help us understand the behavior of points and functions within a 'local' environment. They are crucial when discussing continuity, limits, and, as seen in the exercise, the nature of boundary and isolated points. When we talk about boundary points, one of the first questions is whether any neighborhood of that point contains elements from both the set and its complement鈥攕tressing the role neighborhoods play in real analysis.

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Most popular questions from this chapter

Prove: (a) \(\partial\left(S_{1} \cup S_{2}\right) \subset \partial S_{1} \cup \partial S_{2}\) (b) \(\partial\left(S_{1} \cap S_{2}\right) \subset \partial S_{1} \cup \partial S_{2}\) (c) \(\partial \bar{S} \subset \partial S\) (d) \(\partial S=\partial S^{c}\) (e) \(\partial(S-T) \subset \partial S \cup \partial T\)

Let \(C[0, \infty)\) be the set of all real-valued functions continuous on \([0, \infty)\). For each nonnegative integer \(n,\) let $$ \|f\|_{n}=\max \\{|f(x)| \mid 0 \leq x \leq n\\} $$ and $$ \rho_{n}(f, g)=\frac{\|f-g\|_{n}}{1+\|f-g\|_{n}} $$ Define $$ \rho(f, g)=\sum_{n=1}^{\infty} \frac{1}{2^{n-1}} \rho_{n}(f, g) $$ (a) Show that \(\rho\) is a metric on \(C[0, \infty)\). (b) Let \(\left\\{f_{k}\right\\}_{k=1}^{\infty}\) be a sequence of functions in \(C[0, \infty)\). Show that $$ \lim _{k \rightarrow \infty} f_{k}=f $$ in the sense of Definition 8.1 .14 if and only if $$ \lim _{k \rightarrow \infty} f_{k}(x)=f(x) $$ uniformly on every finite subinterval of \([0, \infty)\). (c) Show that \((C[0, \infty), \rho)\) is complete.

Show that if \(A\) is an arbitrary nonempty set, then $$ \rho(u, v)=\left\\{\begin{array}{ll} 0 & \text { if } v=u \\ 1 & \text { if } v \neq u \end{array}\right. $$ is a metric on \(A\).

Suppose that \(\mathbf{X}=\left\\{x_{i}\right\\}_{i=1}^{\infty}\) is in \(\ell_{p},\) where \(p>1\). Show that (a) \(\mathbf{X} \in \ell_{r}\) for all \(r>p\) (b) If \(r>p,\) then \(\|\mathbf{X}\|_{r} \leq\|\mathbf{X}\|_{p}\) (c) \(\lim _{r \rightarrow \infty}\|\mathbf{X}\|_{r}=\|\mathbf{X}\|_{\infty}\).

Let \(I[a, b]\) be the set of all real-valued functions that are Riemann integrable on \([a, b],\) with \(\rho(u, v)=\sup _{a \leq x \leq b}|u(x)-v(x)| .\) Show that \(f(u)=\int_{a}^{b} u(x) d x\) is a uniformly continuous function from \(I[a, b]\) to \(\mathbb{R}\).
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