Please follow the steps to prove the statement and also show all steps clearly. Thanks for your help ### Solution to Problem 2b: Compactness and Closed Sets**Objective:** Show that if $ K \subseteq \mathbb{R} $ is compact, then $ K $ is closed.**Hint:** Follow these steps to demonstrate the solution.1. **Demonstrate why proving $\mathbb{R} \setminus K$ is open is sufficient:** - Since the complement of $ K $ in $\mathbb{R}$ is open, it follows from the definition of closed sets in topological spaces that $ K $ itself is closed.2. **Prove that $\mathbb{R} \setminus K$ is open using the following outline:** - **Step 1:** Select $ x \in \mathbb{R} \setminus K $ and $ y \in K $ such that $ x \neq y $. - **Step 2:** Establish that there exists $\epsilon_y > 0$ ensuring: \[ D(x, \epsilon_y) \cap D(y, \epsilon_y) = \emptyset \] where $ D(a, \epsilon) $ denotes the open disk centered at $ a $ with radius $\epsilon$. - **Step 3:** Prove that the collection $\mathcal{A} = \{D(y, \epsilon_y)\}_{y \in K}$ forms an open cover of $ K $. - **Step 4:** Given the compactness of $ K $, show the existence of a finite subcover $\{D(y_i, \epsilon_{y_i}) \mid i = 1, 2, \ldots, k\}$ from $\mathcal{A}$. - **Step 5:** Define $\epsilon = \min\{\epsilon_{y_i} \mid i = 1, 2, \ldots, k\} > 0$. - **Step 6:** Show $ D(y_i, \epsilon_{y_i}) \cap D(x, \epsilon) = \emptyset $ for each $ i = 1, 2, \ldots, k $. - **Step 7:** Note that: \[ K \subseteq \bigcup_{y \in K

Given that is compact. we have to show that is closed .In other words ,it is suffices to show that…

Answered: 2b. Show that if K CR is compact then K…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Please follow the steps to prove the statement and also show all steps clearly. Thanks for your help

### Solution to Problem 2b: Compactness and Closed Sets

**Objective:** Show that if $ K \subseteq \mathbb{R} $ is compact, then $ K $ is closed.

**Hint:** Follow these steps to demonstrate the solution.

1. **Demonstrate why proving $\mathbb{R} \setminus K$ is open is sufficient:**

- Since the complement of $ K $ in $\mathbb{R}$ is open, it follows from the definition of closed sets in topological spaces that $ K $ itself is closed.

2. **Prove that $\mathbb{R} \setminus K$ is open using the following outline:**

- **Step 1:** Select $ x \in \mathbb{R} \setminus K $ and $ y \in K $ such that $ x \neq y $.

- **Step 2:** Establish that there exists $\epsilon_y > 0$ ensuring:
\[
D(x, \epsilon_y) \cap D(y, \epsilon_y) = \emptyset
\]
where $ D(a, \epsilon) $ denotes the open disk centered at $ a $ with radius $\epsilon$.

- **Step 3:** Prove that the collection $\mathcal{A} = \{D(y, \epsilon_y)\}_{y \in K}$ forms an open cover of $ K $.

- **Step 4:** Given the compactness of $ K $, show the existence of a finite subcover $\{D(y_i, \epsilon_{y_i}) \mid i = 1, 2, \ldots, k\}$ from $\mathcal{A}$.

- **Step 5:** Define $\epsilon = \min\{\epsilon_{y_i} \mid i = 1, 2, \ldots, k\} > 0$.

- **Step 6:** Show $ D(y_i, \epsilon_{y_i}) \cap D(x, \epsilon) = \emptyset $ for each $ i = 1, 2, \ldots, k $.

- **Step 7:** Note that:
\[
K \subseteq \bigcup_{y \in K

Expert Solution

Step 1: compact implies closed :Proof

Given that $K subset of or equal to straight real numbers$ is compact. we have to show that $K$ is closed .In other words ,it is suffices to show that $straight real numbers set minus K$ is open.

because, $K space i s space c l o s e d space left right double arrow space o u t s i d e space p o i n t space o f space K space i s space n o t space l i m i t space p o i n t space o f space K$

$left right double arrow t h e r e space e x i s t space a t l e a s t space n e i g h b o u r h o o d space o f space e a c h space o u t s i d e space p o i n t space o f space K space w h i c h space i s space d i s j o i n t space f r o m space K$

$left right double arrow f o r space e a c h space k element of K comma there exists space delta subscript k greater than 0 space space s u c h space t h a t space k element of N subscript delta subscript k end subscript open parentheses k close parentheses subset of or equal to straight real numbers set minus K$

$left right double arrow f o r space e a c h space k element of K comma space k space i s space i n t e r i o r space p o i n t space o f space straight real numbers set minus K.$

$left right double arrow straight real numbers set minus K space i s space o p e n$ .

Thus we have to prove that $straight real numbers set minus K$ is open.

for this let $x space element of straight real numbers set minus K$ is arbitrary but fixed point. we have to prove that x is interior point of $straight real numbers set minus K$ .

Now for each $y element of K$ , we have $x not equal to y$

Now $straight real numbers$ is Housdorff space .so from Housdorff propperty of $straight real numbers comma$ for every $y element of K comma there exists space epsilon subscript x y end subscript greater than 0 space s u c h space t h a t space B open parentheses y comma epsilon subscript x y end subscript close parentheses intersection B open parentheses x comma epsilon subscript x y end subscript close parentheses equals ϕ$

where $B open parentheses a comma delta close parentheses space$ mean open open ball with centre at a and radius $delta greater than 0$ .