# Convergence of a Sequence in a Finer Topology

• Bashyboy
In summary, the conversation discusses the convergence of a sequence in a space with different topologies. It is established that a sequence will still converge in a coarser topology, but it may not converge in a finer topology. The conversation also addresses confusion about the proper negation of the definition of convergence, and attempts to find a counterexample to settle the matter. The example of the sequence 1/n in the indiscrete topology is suggested as a potential counterexample.
Bashyboy

## 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.

## 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.

Bashyboy said:
(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##
Neither of these are right.
xn does not converge to x iff there exists an open set U containing x such that for every N>0, there exists n>N with xn not in U.
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.
This is a good example to work with (sort of trivial, but that is ok for an example). Find a sequence that doesn't converge in the standard topology and show that it converges in the indiscrete topology.

## What is convergence of a sequence in a finer topology?

Convergence of a sequence in a finer topology refers to the idea that if a sequence of points in a topological space converges to a certain point in a coarser topology, then it also converges to that same point in a finer topology.

## How does convergence in a finer topology relate to convergence in a coarser topology?

Convergence in a finer topology is a stronger condition than convergence in a coarser topology. This means that if a sequence converges in a finer topology, it will also converge in a coarser topology, but the reverse is not necessarily true.

## What is the significance of convergence in a finer topology?

Convergence in a finer topology is important in understanding the behavior of a sequence in a topological space. It allows us to examine how the sequence behaves as we refine the topology, and can provide insights into the structure and properties of the space.

## How is convergence in a finer topology proven?

To prove convergence of a sequence in a finer topology, one must show that for any open set containing the limit point, there exists a point in the sequence that is also contained in that open set. This ensures that the sequence gets arbitrarily close to the limit point in the finer topology.

## Can a sequence converge in one topology but not in a finer topology?

Yes, it is possible for a sequence to converge in one topology but not in a finer topology. This is because the finer topology may have more open sets, leading to stricter conditions for convergence. However, if a sequence does converge in a finer topology, it will also converge in the coarser topology.

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