Theorem 8.6. Let C be a connected subset of the topological space X. If D is a subset of X such that C CDCC, then D is connected.

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Could you explain how to show 8.6 in detail?

**Theorem 8.6.** Let \( C \) be a connected subset of the topological space \( X \). If \( D \) is a subset of \( X \) such that \( C \subseteq D \subseteq \overline{C} \), then \( D \) is connected.

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**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 \( \overline{A} \cap B = A \cap \overline{B} = \emptyset \). 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.

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

1. \( X \) is connected.

2. There is no continuous function \( f : X \rightarrow \mathbb{R}_{\text{std}} \) 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 exists a finite number of the \( U_\alpha \)’s, \( \{U_{\alpha_1}, U_{\alpha_2}, U_{\alpha_3}, ..., U_{\alpha_n}\} \) such that \( p \in U_{\alpha_1} \), \( q \in U_{\alpha_n} \), and for each \( i < n \
Transcribed Image Text:**Theorem 8.6.** Let \( C \) be a connected subset of the topological space \( X \). If \( D \) is a subset of \( X \) such that \( C \subseteq D \subseteq \overline{C} \), then \( D \) is connected. --- **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 \( \overline{A} \cap B = A \cap \overline{B} = \emptyset \). 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 \( f : X \rightarrow \mathbb{R}_{\text{std}} \) 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 exists a finite number of the \( U_\alpha \)’s, \( \{U_{\alpha_1}, U_{\alpha_2}, U_{\alpha_3}, ..., U_{\alpha_n}\} \) such that \( p \in U_{\alpha_1} \), \( q \in U_{\alpha_n} \), and for each \( i < n \
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