Could you explanin how to prove 2.31 using definitions below? **Theorem 2.30.** Let $ A $ be a subset of the topological space $ X $, and let $ p $ be a point in $ X $. If the set $\{x_i\}_{i \in \mathbb{N}} \subset A$ and $ x_i \to p $, then $ p $ is in the closure of $ A $.As we shall see later, in some topological spaces, the converse of the previous result is not true. But it is true for $\mathbb{R}^n$.**Theorem 2.31.** In the standard topology on $\mathbb{R}^n$, if $ p $ is a limit point of a set $ A $, then there is a sequence of points in $ A $ that converges to $ p $.**Definition.** Let $ (X, \mathcal{J}) $ be a topological space, $ A $ a subset of $ X $, and $ p $ a point in $ X $. Then $ p $ is a *limit point* of $ A $ if and only if for each open set $ U $ containing $ p $, $ (U - \{p\}) \cap A \neq \emptyset $. Notice that $ p $ may or may not belong to $ A $.**Definition.** A *sequence* in a topological space $ X $ is a function from $ \mathbb{N} $ to $ X $. The image of $ i $ under this function is a point of $ X $ denoted $ x_i $ and we traditionally write the sequence by listing its images: $ x_1, x_2, x_3, \ldots $ or in shorter form: $(x_i)_{i \in \mathbb{N}}$.**Definition.** A point $ p \in X $ is a *limit of the sequence* $(x_i)_{i \in \mathbb{N}}$, or, equivalently, $(x_i)_{i \in \mathbb{N}}$ *converges to* $ p $ (written $ x_i \to p $), if and only if for every open set $ U $

now suppose that p is a limit point of A .then for each n∈N we have 1n>0,hence there exists some…

Answered: Theorem 2.31. In the standard topology…

Theorem 2.31. In the standard topology on R", if p is a limit point of a set A, then there is a sequence of points in A that converges to p.

Advanced Engineering Mathematics

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ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

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Could you explanin how to prove 2.31 using definitions below?

**Theorem 2.30.** Let $ A $ be a subset of the topological space $ X $, and let $ p $ be a point in $ X $. If the set $\{x_i\}_{i \in \mathbb{N}} \subset A$ and $ x_i \to p $, then $ p $ is in the closure of $ A $.

As we shall see later, in some topological spaces, the converse of the previous result is not true. But it is true for $\mathbb{R}^n$.

**Theorem 2.31.** In the standard topology on $\mathbb{R}^n$, if $ p $ is a limit point of a set $ A $, then there is a sequence of points in $ A $ that converges to $ p $.

**Definition.** Let $ (X, \mathcal{J}) $ be a topological space, $ A $ a subset of $ X $, and $ p $ a point in $ X $. Then $ p $ is a *limit point* of $ A $ if and only if for each open set $ U $ containing $ p $, $ (U - \{p\}) \cap A \neq \emptyset $. Notice that $ p $ may or may not belong to $ A $.

**Definition.** A *sequence* in a topological space $ X $ is a function from $ \mathbb{N} $ to $ X $. The image of $ i $ under this function is a point of $ X $ denoted $ x_i $ and we traditionally write the sequence by listing its images: $ x_1, x_2, x_3, \ldots $ or in shorter form: $(x_i)_{i \in \mathbb{N}}$.

**Definition.** A point $ p \in X $ is a *limit of the sequence* $(x_i)_{i \in \mathbb{N}}$, or, equivalently, $(x_i)_{i \in \mathbb{N}}$ *converges to* $ p $ (written $ x_i \to p $), if and only if for every open set $ U $

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