Sequences and Nets: Does (1/n) Converge to [0,1]?

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Discussion Overview

The discussion revolves around the convergence of the sequence (1/n) within a specific topology on the real numbers, particularly focusing on the neighborhood properties of the interval [0,1]. Participants explore whether (1/n) converges to every point in [0,1] under the proposed topological structure.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant suggests that if the real numbers are given a topology \left\{R,\phi, [0,1]\right\}, then (1/n) converges to every point of [0,1] since it is a neighborhood of every point.
  • Another participant questions the definitions of R and \phi, asking whether the proposed topology satisfies the definition of a topology.
  • A different participant asserts that if \mathbb{R} has the topology \{\emptyset,[0,1],\mathbb{R}\}, then (1/n) converges to every point of \mathbb{R}.
  • A later reply acknowledges the previous point about convergence in the specified topology, indicating some level of agreement.

Areas of Agreement / Disagreement

Participants express differing views on the validity of the proposed topology and its implications for convergence. There is no consensus on the correctness of the topology or the convergence claims.

Contextual Notes

Limitations include potential ambiguities in the definitions of the topology and the sequence's convergence properties, as well as the need for clarification on the topology's adherence to formal definitions.

ForMyThunder
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Say the real numbers were given a topology \left\{R,\phi, [0,1]\right\}. Does the sequence (1/n) converge to every point of [0,1] since it is a neighborhood of every point?
 
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What is R and \phi? Does your topology satisfy the definition of a topology?
 
If \mathbb{R} has the topology \{\emptyset,[0,1],\mathbb{R}\}, then the sequence (1/n) converges to every point of \mathbb{R}!
 
micromass said:
If \mathbb{R} has the topology \{\emptyset,[0,1],\mathbb{R}\}, then the sequence (1/n) converges to every point of \mathbb{R}!

Yeah, I guess you're right. Thanks.
 

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