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Chapter 5: Problem 14

Prove: If \(A\) and \(B\) are open and \(A \cap B=\emptyset\), then \(A \cap \bar{B}=\bar{A} \cap B=\emptyset\).

Short Answer

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Question: Prove that if \(A\) and \(B\) are open sets with no common elements, their intersections with the closures of each other are also empty sets, i.e., \(A \cap \bar{B} = \emptyset\) and \(\bar{A} \cap B = \emptyset\). Answer: We proved that \(A \cap \bar{B} = \emptyset\) and \(\bar{A} \cap B = \emptyset\) by contradiction. For each case, we assumed there exists an element in the intersection, and by examining the properties of open sets, closure, and limit points, we reached a contradiction, concluding that the intersections must be empty sets.

Step by step solution

01

Recall Definitions

Let's recall the definition of open sets, intersection of sets, and closure of a set. - Open Set: A set \(U\) is called an open set if for every point \(x\) in \(U\), there exists an open ball (in metric space) or neighborhood (in topological space) centered at \(x\) that is entirely contained within \(U\). - Intersection of Sets: The intersection of two sets \(A\) and \(B\), denoted by \(A \cap B\), is the set containing all the elements that belong to both \(A\) and \(B\). - Closure of a Set: The closure of a set \(A\), denoted by \(\bar{A}\), is the smallest closed set containing \(A\). A closed set is the complement of an open set, which means that it contains all its limit points
02

Prove \(A \cap \bar{B} = \emptyset\) by contradiction

Let's assume that there exists an element \(x\) which belongs to both \(A\) and \(\bar{B}\). By definition of closure, this means that \(x\) either belongs to \(B\) or is a limit point of \(B\). Since we know that \(A \cap B = \emptyset\), \(x\) cannot belong to \(B\) (otherwise it would be a common element of \(A\) and \(B\)). That means \(x\) must be a limit point of \(B\). However, \(A\) is an open set, so there must exist an open neighborhood (or open ball) \(N\) centered at \(x\) that is entirely within \(A\). Since \(x\) is also a limit point of \(B\), there must be a point \(y\) in the intersection of \(N\) and \(B\), i.e., \(y \in N \cap B\). But since \(N \subseteq A\) and \(y \in N\), \(y\) is also in \(A\). This is a contradiction because we initially assumed that \(A \cap B = \emptyset\). Therefore, our assumption that there exists an element \(x\) in \(A \cap \bar{B}\) must be incorrect, and we can conclude that \(A \cap \bar{B} = \emptyset\).
03

Prove \(\bar{A} \cap B = \emptyset\) by contradiction

Similar to Step 2, we will assume there exists an element \(x\) which belongs to both \(\bar{A}\) and \(B\). By definition of closure, this means that \(x\) either belongs to \(A\) or is a limit point of \(A\). Since we know that \(A \cap B = \emptyset\), \(x\) cannot belong to \(A\). That means \(x\) must be a limit point of \(A\). However, \(B\) is an open set, so there must exist an open neighborhood (or open ball) \(N\) centered at \(x\) that is entirely within \(B\). Since \(x\) is also a limit point of \(A\), there must be a point \(y\) in the intersection of \(N\) and \(A\), i.e., \(y \in N \cap A\). But since \(N \subseteq B\) and \(y \in N\), \(y\) is also in \(B\). This is a contradiction because we initially assumed that \(A \cap B = \emptyset\). Therefore, our assumption that there exists an element \(x\) in \(\bar{A} \cap B\) must be incorrect, and we can conclude that \(\bar{A} \cap B = \emptyset\). We have shown that \(A \cap \bar{B} = \emptyset\) and \(\bar{A} \cap B = \emptyset\), completing the proof.

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Key Concepts

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

Open Sets
In topology, an open set is fundamental for understanding the structure and "neighborhoods" within a space. Imagine an open set as an area where each point has some "space" around it that is still within the set. It's like standing anywhere in a park and knowing you can move a little in any direction without leaving the park.
Every point in an open set has an open neighborhood around it. In a metric space with a defined distance, this neighborhood is an open ball centered at the point.
  • An open set does not include its boundary. If you think about the surface of a bubble, any point on the surface can move inwards back into the bubble, not touching the air outside.
  • Sets like intervals (open intervals like (a, b)) are open since you can pick any fraction between a and b and move slightly left or right without leaving the interval.
Understanding open sets helps in dealing with continuity, limits, and many other important concepts in mathematics.
Intersection of Sets
The intersection of sets combines elements that are common to multiple sets. When you say "intersection," think of it as a shared pathway or mutual ground. Symbolically, the intersection of two sets, let's say, \(A\) and \(B\), is written as \(A \cap B\). This intersection includes only those elements that belong to both sets.
  • If \(A \cap B = \emptyset\), it means there are no common elements, and we call \(A\) and \(B\) disjoint.
  • Intersections help us understand related concepts like shared characteristics or overlapping properties among datasets.
In essence, the intersection of sets is like finding the common friends in two different friend groups, showing only those people who appear in both circles.
Closure of a Set
A closure of a set \(A\), denoted by \(\bar{A}\), is like sealing all possible "leaks" of \(A\) to include limit points where sequences may approach the set but do not quite fall into \(A\). These limit points are crucial because they tell you about the behavior of \(A\) in conjunction with its surrounding space.
Let's break this down:
  • The closure includes all points in \(A\) and any limit point that can't be separated with an open set. It's like an invisible boundary ensuring every path leads only to the closure without an exit route.
  • Closed sets contain all their limit points. Thus, a set and its closure are identically closed if \(\bar{A}\) equals \(A\).
Think of closure as "completing" a set, ensuring it captures every point that can "almost make it" into \(A\) based on the surrounding topology and its properties.

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Most popular questions from this chapter

Find all first-order partial derivatives. (a) \(f(x, y, z)=\log (x+y+2 z)\) (b) \(f(x, y, z)=x^{2}+3 x y z+2 x y\) (c) \(f(x, y, z)=x e^{y z}\) (d) \(f(x, y, z)=z+\sin x^{2} y\)

Determine whether \(\lim _{\mathbf{X} \rightarrow \mathbf{X}_{0}} f(\mathbf{X})=\pm \infty\). (a) \(f(\mathbf{X})=\frac{|\sin (x+2 y+4 z)|}{(x+2 y+4 z)^{2}}, \quad \mathbf{X}_{0}=(2,-1,0)\) (b) \(f(\mathbf{X})=\frac{1}{\sqrt{x-y}}, \quad \mathbf{X}_{0}=(0,0)\) (c) \(f(\mathbf{X})=\frac{\sin 1 / x}{\sqrt{x-y}}, \quad \mathbf{X}_{0}=(0,0)\) (d) \(f(\mathbf{X})=\frac{4 y^{2}-x^{2}}{(x-2 y)^{3}}, \quad \mathbf{X}_{0}=(2,1)\) (e) \(f(\mathbf{X})=\frac{\sin (x+2 y+4 z)}{(x+2 y+4 z)^{2}}, \quad \mathbf{X}_{0}=(2,-1,0)\)

Prove that \(d(S)=d(\bar{S})\) for any set \(S\) in \(\mathbb{R}^{n}\).

Suppose that \(u, v,\) and \(f\) are defined on \((-\infty, \infty)\). Let \(u\) and \(v\) be differentiable and \(f\) be continuous for all \(x\). Show that $$ \frac{d}{d x} \int_{u(x)}^{v(x)} f(t) d t=f(v(x)) v^{\prime}(x)-f(u(x)) u^{\prime}(x) . $$

Suppose that \(p=p\left(x_{1}, x_{2}, \ldots, x_{n}\right)\) is a homogeneous polynomial of degree \(r\) (Exercise 5.4.8). Let \(i_{1}, i_{2}, \ldots, i_{n}\) be nonnegative integers such that $$ i_{1}+i_{2}+\cdots+i_{n}=k $$ and let $$ q\left(x_{1}, x_{2}, \ldots, x_{n}\right)=\frac{\partial^{k} p\left(x_{1}, x_{2}, \ldots, x_{n}\right)}{\partial x_{1}^{i_{1}} \partial x_{2}^{i_{2}} \cdots \partial x_{n}^{i_{n}}} $$ Show that \(q\) is homogeneous of degree \(\leq r-k,\) subject to the convention that a homogeneous polynomial of negative degree is identically zero.
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