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Sequence Limits are unique in Hausdorff Spaces

Last updated Nov 1, 2022

# Statement

Suppose (X,τ)(X, \tau) is a Hausdorff

Hausdorff

Definition Let (X,τ)(X, \tau) be a . Then XX is if for every x,yXx, y \in X there exist $U,...

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. Let (xn)X(x_n) \subset X. Then if xnxXx_{n} \to x \in X and xnyYx_{n} \to y \in Y, x=yx=y.

# Proof 1

Suppose not. Then there exists a sequence (xn)X(x_n) \subset X so that xnxx_{n} \to x and xnyx_{n} \to y and xyx \neq y. Since XX is a Hausdorff

Hausdorff

Definition Let (X,τ)(X, \tau) be a . Then XX is if for every x,yXx, y \in X there exist $U,...

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, there exists U,VXU, V \subset X Open

Open

Definition Suppose (X,τ)(X, \tau) is a . Then UXU \subset X is if UτU \in \tau....

11/7/2022

so that xUx \in U, yVy \in V, and UV=U \cap V = \emptyset. Since xnxx_{n} \to x, there exists NNN \in \mathbb{N} so that nN\forall n \geq N, we have that xnUx_{n} \in U. Likewise, since xnyx_{n} \to y, there exists MNM \in \mathbb{N} so that mM\forall m \geq M, we have that xmVx_{m} \in V. Then UVxnn=max(M,N)U \cap V \supset {x_n}_{n=\max(M, N)}^{\infty}

Contradicting

the Mutually Disjoint

Mutually Disjoint

Definition Let {Ai}iI\{A{i}\}{i \in I} be a collectio of s indexe by II. We say {Ai}iI\{A{i}\}{i \in I} is if...

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ness of UU and VV. \blacksquare

# Proof 2

Net Limits are unique in Hausdorff Spaces

Net Limits are unique in Hausdorff Spaces

Statement Let XX be and let (xα)αAX(x{\alpha}){\alpha \in A} \subset 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 XX be a and let {xn}n=1X\{xn\}{n=1}^{\infty} \subset X be a . Then {xn}n=1\{xn\}{n=1}^{\infty} is a . Proof This follows by...

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. \square