How do I show (4),(5),(6) in 8.1? **Definition.** Let $ X $ be a topological space. Then $ X $ is **connected** if and only if $ X $ is not the union of two disjoint non-empty open sets.**Definition.** Let $ X $ be a topological space. Subsets $ A, B $ in $ X $ are **separated** if and only if $ \overline{A} \cap B = A \cap \overline{B} = \varnothing $. Thus $ B $ does not contain any limit points of $ A $, and $ A $ does not contain any limit points of $ B $. The notation $ X = A \mid B $ means $ X = A \cup B $ and $ A $ and $ B $ are separated sets.**Theorem 8.1.** The following are equivalent:1. $ X $ is connected.2. There is no continuous function $ f : X \to \mathbb{R}_{std} $ such that $ f(X) = \{0, 1\} $.3. $ X $ is not the union of two disjoint non-empty separated sets.4. $ X $ is not the union of two disjoint non-empty closed sets.5. The only subsets of $ X $ that are both closed and open in $ X $ are the empty set and $ X $ itself.6. For every pair of points $ p $ and $ q $ and every open cover $\{U_\alpha\}_{\alpha \in \Lambda} $ of $ X $ there exist a finite number of the $ U_\alpha $'s, $\{U_{\alpha_1}, U_{\alpha_2}, U_{\alpha_3}, \ldots, U_{\alpha_n}\}$ such that $ p \in U_{\alpha_1}, q \in U_{\alpha_n} $, and for each $ i < n, U_{\alpha_i} \cap U_{\alpha_{i+1}} \neq \varnothing $.

as given all the conditions are equivalent (4) To show - X is not the union of two disjoint non empty closed sets. given (3) X is not the union of two disjoint non empty separated sets. we will show 3→4 on the contrary , we suppose that X is the union of two disjoint non empty closed sets. that is , X=A∪B where A and B are two non empty disjoint closed sets. then X=A∪B and since both A and B are closed thus A¯=AB¯=B⇒A¯∪B¯=A∪B=ϕ implies that X is separated by two non empty disjoint sets A and B which is contradiction to given statement 3, that is X is not the union of two disjoint non empty separated sets thus our assumption is wrong and we have that X is not the union of two disjoint non empty closed sets (5) to show : the only subset of X that are both closed and open in X are the empty set and X itself proof on the contrary , we suppose that A is a subset of X that is both closed and empty ⇒A is both closed and open this implies X-A is closed in X such that X=A∪X-A and A∩X-A=ϕ this implies X is union of two non empty disjoint sets, which is a contradiction to given condition (4) that is X is not the union of two non empty disjoint sets therefore, our assumption is wrong hence we can say the only subset of X that are both closed and open in X are the empty set and X itself

Math

Advanced Math

(3) X is not the union of two disjoint non-empty separated sets. (4) X is not the union of two disjoint non-empty closed sets. (5) The only subsets of X that are both closed and open in X are the empty set and X itself. (6) For every pair of points p and q and every open cover {Uq}aea of X there exist a finite number of the Uq's, {U«,, Uaz, Uaz»….. , Uan} such that p E Ua,, q E Ugn; and for each i < n, Ua; n Uai+1 + Ø.

(3) X is not the union of two disjoint non-empty separated sets. (4) X is not the union of two disjoint non-empty closed sets. (5) The only subsets of X that are both closed and open in X are the empty set and X itself. (6) For every pair of points p and q and every open cover {Uq}aea of X there exist a finite number of the Uq's, {U«,, Uaz, Uaz»….. , Uan} such that p E Ua,, q E Ugn; and for each i < n, Ua; n Uai+1 + Ø.

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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Concept explainers

Counting Principles

In statistics, the counting principle can be said to be the concept in which the quantity of an item is determined. There can be a single item, or a group of items, or items that are related to each other. In every possible set of items, the quantity of the possible item groups or their sorting measures can be determined with counting principles.

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How do I show (4),(5),(6) in 8.1?

**Definition.** Let $ X $ be a topological space. Then $ X $ is **connected** if and only if $ X $ is not the union of two disjoint non-empty open sets.

**Definition.** Let $ X $ be a topological space. Subsets $ A, B $ in $ X $ are **separated** if and only if $ \overline{A} \cap B = A \cap \overline{B} = \varnothing $. Thus $ B $ does not contain any limit points of $ A $, and $ A $ does not contain any limit points of $ B $. The notation $ X = A \mid B $ means $ X = A \cup B $ and $ A $ and $ B $ are separated sets.

**Theorem 8.1.** The following are equivalent:

1. $ X $ is connected.

2. There is no continuous function $ f : X \to \mathbb{R}_{std} $ such that $ f(X) = \{0, 1\} $.

3. $ X $ is not the union of two disjoint non-empty separated sets.

4. $ X $ is not the union of two disjoint non-empty closed sets.

5. The only subsets of $ X $ that are both closed and open in $ X $ are the empty set and $ X $ itself.

6. For every pair of points $ p $ and $ q $ and every open cover $\{U_\alpha\}_{\alpha \in \Lambda} $ of $ X $ there exist a finite number of the $ U_\alpha $'s, $\{U_{\alpha_1}, U_{\alpha_2}, U_{\alpha_3}, \ldots, U_{\alpha_n}\}$ such that $ p \in U_{\alpha_1}, q \in U_{\alpha_n} $, and for each $ i < n, U_{\alpha_i} \cap U_{\alpha_{i+1}} \neq \varnothing $.

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