Question

Prove: If a nonempty subset \(S\) of \(\mathbb{R}^{n}\) is both open and closed, then \(S=\mathbb{R}^{n}\).

Short answer

Question: Prove that any nonempty open and closed subset must be equal to \(\mathbb{R}^n\). Answer: Suppose that \(S \subset \mathbb{R}^n\) is nonempty, open, and closed, and that \(S \neq \mathbb{R}^n\). Let \(x \in \mathbb{R}^n\) be a point not in \(S\), and let \(S'\) be the complement of \(S\) in \(\mathbb{R}^n\). Since \(S\) is closed, its complement \(S'\) is open. Now define a function \(f: S \to S'\) such that \(f(p) = x\). Consider the distance \(\|p - f(p)\|= \Vert p - x \Vert\) and find a point \(p_0 \in S\) where this distance achieves its infimum. This infimum distance is denoted by \(d\). Since \(S\) and \(S'\) are both open, there exist open balls \(B_{r_1}(p_0) \subset S\) and \(B_{r_2}(x) \subset S'\) with positive radii \(r_1\) and \(r_2\). Let \(r = \min\{ r_1, r_2\}\). Consider the point \(p_1 = p_0 + \frac{r}{2}\frac{x-p_0}{\Vert x-p_0\Vert}\). Observe that \(p_1\) belongs both to \(B_r(p_0) \subset S\) and \(B_{r_2}(x) \subset S'\), which contradicts the disjointness of \(S\) and \(S'\). This contradiction implies that our initial assumption is incorrect, and thus \(S = \mathbb{R}^n\).

Step-by-step solution

Define the complement set and assume there exists a point in it

Let's assume \(S \neq \mathbb{R}^n\). Then there must be a point \(x \in \mathbb{R}^n\) such that \(x \notin S\). Let \(S'\) denote the complement of \(S\) in \(\mathbb{R}^n\), i.e., \(S' = \mathbb{R}^n \setminus S\). Since \(x \notin S\), we have \(x \in S'\).

Prove that \(S'\) is open

As \(S\) is closed, by definition, its complement \(\mathbb{R}^n \setminus S = S'\) must be open. Since \(S'\) is open, for each point \(y \in S'\), there exists a positive radius \(r_y\) such that the open ball \(B_{r_y}(y)\) centered at \(y\) and with radius \(r_y\) is contained in \(S'\).

Show that \(S\) and \(S'\) are disjoint and cover \(\mathbb{R}^n\)

Let's note that, by construction, \(S\) and \(S'\) are disjoint, i.e., \(S \cap S' = \emptyset\). Moreover, \(S \cup S' = \mathbb{R}^n\), since every point in \(\mathbb{R}^n\) belongs to either the set \(S\) or its complement \(S'\).

Define a function from \(S\) to \(S'\)

Now, let's define a function \(f: S \to S'\) such that \(f(p) = x\), where \(x\) is the point in \(S'\) as defined in Step 1. Since \(x \in S'\) and \(S'\) is open, \(f(p) = x\) is defined for every \(p \in S\).

Show that there must be a closest point to \(f(p)\) in \(S\)

Let's consider the distance \(\|p - f(p)\|= \Vert p - x \Vert\) between the point \(p \in S\) and its corresponding image point \(f(p) \in S'\). Since \(S\) is nonempty, it is possible to find the point \(p_0 \in S\) such that this distance achieves its infimum: $$ p_0 = \text{argmin}_{p \in S} \Vert p - x \Vert . $$ Let's denote this minimum distance as \(d = \Vert p_0 - x\Vert\), i.e., $$ d = \inf_{p \in S} \Vert p - x \Vert . $$

Find a contradiction with the open property of \(S\) and \(S'\)

Now comes a critical point in our proof. Since both \(S\) and \(S'\) are open, there exist open balls \(B_{r_1}(p_0) \subset S\) and \(B_{r_2}(x) \subset S'\) with positive radii \(r_1\) and \(r_2\). Let \(r = \min\{ r_1, r_2\}>0\). Then, we have the open ball \(B_r(p_0) \subset S\). Moreover, the open ball \(B_r(x) \subset S'\). However, the point \(p_1 = p_0 + \frac{r}{2}\frac{x-p_0}{\Vert x-p_0\Vert} \in B_r(p_0) \subset S\) since \(\Vert p_1 - p_0 \Vert = \frac{r}{2} < r \leq r_1\). On the other hand, observe that \(\Vert p_1 - x \Vert = \frac{d + r}{2} < \frac{d + r_1}{2} \leq r_2\), which implies that \(p_1 \in B_{r_2}(x) \subset S'\). But this means \(p_1\) lies both in \(S\) and \(S'\), which contradicts the fact that \(S\) and \(S'\) are disjoint. This contradiction shows that our assumption that \(S \neq \mathbb{R}^n\) is wrong. Therefore, we conclude that \(S = \mathbb{R}^n\).

Key concepts

Open and Closed Sets in Real Analysis

In real analysis, understanding the properties of open and closed sets is fundamental. An open set is a set where, for every point within the set, there is some neighborhood around that point that is entirely contained within the set. This neighborhood can be visualized as an open ball where all its points are still within the set, hence the term 'open.'

Conversely, a closed set includes its boundary. In other words, it contains all its limit points, which means that any convergent sequence within the set has its limit also within the set. The complement of an open set is closed, and vice versa. In the Euclidean space, any set can be broadly classified as open, closed, both (clopen), or neither.

Intuitively, if we think of a set as a region in space, open sets can expand infinitely small in any direction within the set without leaving it, and closed sets contain every point right up to their edges. This concept is pivotal in the given exercise, where a nonempty subset of \(\mathbb{R}^n\) that is both open and closed must be the whole space itself, as proven by contradiction in the exercise solution.

Complement of a Set

The complement of a set is a fundamental concept in set theory and real analysis. The complement of a set \(S\) within a given universal set \(U\) includes all the elements that are in \(U\) but not in \(S\). It is denoted as \(U \setminus S\) or simply \(S'\) when the context of \(U\) is clear. In the context of real analysis, when we talk about \(\mathbb{R}^n\), the complement of a set contains all other points in that n-dimensional space that aren't part of the set.

Properties of complements are vital in understanding open and closed sets, as mentioned above, since the complement of an open set is closed and vice versa. The exercise uses the concept of set complements to show that if there's a point not in the subset \(S\), then it's in the complement of \(S\), designated \(S'\). This understanding paves the way to establishing that \(S\) and \(S'\) cover the entire space and therefore must be closely examined to solve the given proof.

Distance in Euclidean Space

The concept of distance in Euclidean space is essential for analyzing the relationship between points in real analysis. The distance between two points in an n-dimensional Euclidean space (\(\mathbb{R}^n\)) is given by the familiar Euclidean distance. For points \(p\) and \(q\), the distance is defined as \(\Vert p - q \Vert = \sqrt{(p_1 - q_1)^2 + \ldots + (p_n - q_n)^2}\), where \(p_1, \ldots, p_n\) and \(q_1, \ldots, q_n\) are the components of the two points.

This distance measure forms the backbone of the

Understanding the Distance and Open Balls

The concept of distance in Euclidean space is crucial to many principles in real analysis, including the characterization of open and closed sets. The distance between any two points, say \(p\) and \(q\), in a Euclidean space \(\mathbb{R}^n\), is determined by the Euclidean distance formula: \(\|p - q\| = \sqrt{(p_1 - q_1)^2 + \ldots + (p_n - q_n)^2}\), where \(p_1, \ldots, p_n\) and \(q_1, \ldots, q_n\) are the coordinates of \(p\) and \(q\) respectively.

An open ball, denoted as \(B_r(x)\), centered at point \(x\) with radius \(r\), is the set of points in \(\mathbb{R}^n\) whose distance from \(x\) is less than \(r\). It's a fundamental building block for defining open sets, which are simply unions of open balls. In the exercise, the concept of distance is used to find a contradiction by showing a point that should be in one set by the distance property ends up being in both of the disjoint sets, \(S\) and its complement, which leads to the conclusion of the proof.