Could you explain how to show 8.9 in detail? **Theorem 8.9.** Let $ f : X \rightarrow Y $ be a continuous, surjective function. If $ X $ is connected, then $ Y $ 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 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 \emptyset $.

Name: Could you explain how to show 8.9 in detail? **Theorem 8.9.** Let $ f
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Description: Could you explain how to show 8.9 in detail? **Theorem 8.9.** Let \( f : X \rightarrow Y $ be a continuous, surjective function. If $ X $ is connected, then $ Y $ is connected.**Definition.** Let $ X $ be a topological space. Then $ X $ is *

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Answered: Theorem 8.9. Let f : X Y be a…

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Theorem 8.9. Let f : X Y be a continuous, surjective function. If X is connected, then Y is connected.

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Could you explain how to show 8.9 in detail?

**Theorem 8.9.** Let $ f : X \rightarrow Y $ be a continuous, surjective function. If $ X $ is connected, then $ Y $ 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 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 \emptyset $.

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