Sequence Limits are unique in Hausdorff Spaces Last updated
Nov 1, 2022
Table of Contents Proof 1 Proof 2 # StatementSuppose ( X , τ ) (X, \tau) ( X , τ ) is a Hausdorff
Hausdorff
Definition
Let ( X , τ ) (X, \tau) ( X , τ ) be a . Then X X X is if for every x , y ∈ X x, y \in X x , y ∈ X there exist $U,...
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. Let ( x n ) ⊂ X (x_n) \subset X ( x n ) ⊂ X . Then if x n → x ∈ X x_{n} \to x \in X x n → x ∈ X and x n → y ∈ Y x_{n} \to y \in Y x n → y ∈ Y , x = y x=y x = y .
# Proof 1Suppose not. Then there exists a sequence ( x n ) ⊂ X (x_n) \subset X ( x n ) ⊂ X so that x n → x x_{n} \to x x n → x and x n → y x_{n} \to y x n → y and x ≠ y x \neq y x = y . Since X X X is a Hausdorff
Hausdorff
Definition
Let ( X , τ ) (X, \tau) ( X , τ ) be a . Then X X X is if for every x , y ∈ X x, y \in X x , y ∈ X there exist $U,...
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, there exists U , V ⊂ X U, V \subset X U , V ⊂ X Open
Open
Definition
Suppose ( X , τ ) (X, \tau) ( X , τ ) is a . Then U ⊂ X U \subset X U ⊂ X is if U ∈ τ U \in \tau U ∈ τ ....
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so that x ∈ U x \in U x ∈ U , y ∈ V y \in V y ∈ V , and U ∩ V = ∅ U \cap V = \emptyset U ∩ V = ∅ . Since x n → x x_{n} \to x x n → x , there exists N ∈ N N \in \mathbb{N} N ∈ N so that ∀ n ≥ N \forall n \geq N ∀ n ≥ N , we have that x n ∈ U x_{n} \in U x n ∈ U . Likewise, since x n → y x_{n} \to y x n → y , there exists M ∈ N M \in \mathbb{N} M ∈ N so that ∀ m ≥ M \forall m \geq M ∀ m ≥ M , we have that x m ∈ V x_{m} \in V x m ∈ V . Then U ∩ V ⊃ x n n = max ( M , N ) ∞ U \cap V \supset {x_n}_{n=\max(M, N)}^{\infty} U ∩ V ⊃ x n n = m a x ( M , N ) ∞
Contradicting
Proof by Contradiction
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the Mutually Disjoint
Mutually Disjoint
Definition
Let { A i } i ∈ I \{A{i}\}{i \in I} { A i } i ∈ I be a collectio of s indexe by I I I . We say { A i } i ∈ I \{A{i}\}{i \in I} { A i } i ∈ I is if...
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ness of U U U and V V V . ■ \blacksquare ■
# Proof 2Net Limits are unique in Hausdorff Spaces
Net Limits are unique in Hausdorff Spaces
Statement
Let X X X be and let ( x α ) α ∈ A ⊂ X (x{\alpha}){\alpha \in A} \subset X ( x α ) α ∈ A ⊂ X be a that converge to both $x,y \in...
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and A Sequence is a Net
A Sequence is a Net
Statement
Let X X X be a and let { x n } n = 1 ∞ ⊂ X \{xn\}{n=1}^{\infty} \subset X { x n } n = 1 ∞ ⊂ X be a . Then { x n } n = 1 ∞ \{xn\}{n=1}^{\infty} { x n } n = 1 ∞ is a .
Proof
This follows by...
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. □ \square □