- #1

Bashyboy

- 1,421

- 5

## Homework Statement

Clearly if a sequence of points ##\{x_n\}## in some space ##X## with some topology, then the sequence will also converge when ##X## is endowed with any coarser topology. I suspect this doesn't hold for endowment of ##X## with a finer topology, since a finer topology amounts to more open sets, decreasing the likelihood of convergence. However, I would like to build a counterexample to settle this matter.

## Homework Equations

## The Attempt at a Solution

First, I am having a embarrassing confusion with quantifiers. The definition of convergence I am working is the following: ##x_n## converges to ##x## if and only if for every open neighborhood ##U## of ##x##, there exists ##N \in \mathbb{N}## such that ##x_n \in U## for all ##n \ge N##. Which of the two is the proper negation:

(1) ##x_n## does not converge to ##x## iff there exists an open neighborhood ##U## of ##x## such that ##x_n \notin U##

**for some**##n \ge N##

(2) ##x_n## does not converge to ##x## iff there exists an open neighborhood ##U## of ##x## such that ##x_n \notin U##

**for every**##n \ge N##

Once this is settled, my goal is to show that the sequence ##\frac{1}{n}## does not converge to ##0## in the lower limit topology (note, I don't actually know if this is the case; but I conjectured it on the basis that it is the simplest example).

EDIT: Perhaps this is easier. Just consider the indiscrete topology on ##\mathbb{R}##. Then every number in ##\mathbb{R}## is a limit of ##\frac{1}{n}##, but the sequence ##\frac{1}{n}## in the standard topology only has ##0## as its limit.