# Convergence of a Sequence in a Finer Topology

1. Jan 5, 2017

### Bashyboy

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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.

2. Jan 5, 2017

### FactChecker

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.
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.