MHB Convergence as for the cofinite topology on R

mathmari
Gold Member
MHB
Messages
4,984
Reaction score
7
Hey! :giggle:

Does the sequence $x_n=\frac{1}{n}$ converges as for the cofinite topology on $\mathbb{R}$ ? If it converges,where does it converge? Could you explain to me what exactly is meant by "cofinite topology on $\mathbb{R}$" ? Do we have to define first this set and then check if we have convergence inside that set? :unsure:
 
Physics news on Phys.org
mathmari said:
Does the sequence $x_n=\frac{1}{n}$ converges as for the cofinite topology on $\mathbb{R}$ ? If it converges,where does it converge?

Could you explain to me what exactly is meant by "cofinite topology on $\mathbb{R}$" ? Do we have to define first this set and then check if we have convergence inside that set?
Hey mathmari!

Wiki defines cofinite topology here.
It's the topology where every open set must either be the empty set, or it must have a finite complement. 🤔

Additionally we need the definition for convergence in a topology. We cannot use the usual $\varepsilon-\delta$ method, since distances are not defined in a topology.
Can we find that definition? 🤔
 
Klaas van Aarsen said:
Wiki defines cofinite topology here.
It's the topology where every open set must either be the empty set, or it must have a finite complement. 🤔

Additionally we need the definition for convergence in a topology. We cannot use the usual $\varepsilon-\delta$ method, since distances are not defined in a topology.
Can we find that definition? 🤔

Do we use the following definition?

$\langle x_n:n\in\mathbb{N}\rangle$ converges to $x$ if and only if for each openneighborhood $U$ of $x$ there is an $m\in\mathbb{N}$ such that $x_n\in U$ whenever $n\ge m_U$.

:unsure:
 
mathmari said:
Do we use the following definition?

$\langle x_n:n\in\mathbb{N}\rangle$ converges to $x$ if and only if for each openneighborhood $U$ of $x$ there is an $m\in\mathbb{N}$ such that $x_n\in U$ whenever $n\ge m_U$.
Yep. (Nod)
 
Klaas van Aarsen said:
Yep. (Nod)

In general it holds that $\frac{1}{n}\rightarrow -$, so do we have to check if for each open neighborhood $U$ of $0$ there is an $m\in \mathbb{N}$ such that $\frac{1}{n}\in U$ whenever $n\geq m_U$ ?
Or do we apply the definition in practice? :unsure:
 
mathmari said:
In general it holds that $\frac{1}{n}\rightarrow -$, so do we have to check if for each open neighborhood $U$ of $0$ there is an $m\in \mathbb{N}$ such that $\frac{1}{n}\in U$ whenever $n\geq m_U$ ?
Yep. (Nod)

What does an open neighborhood of $0$ look like? 🤔
 
Klaas van Aarsen said:
What does an open neighborhood of $0$ look like? 🤔

Is it a ball with center the originand radius $\epsilon>0$ ? :unsure:
 
mathmari said:
Is it a ball with center the originand radius $\epsilon>0$ ? :unsure:
Nope. (Shake)

We can only have a ball with a radius if we can measure distances. But in a topology those are not defined. Instead a neighborhood of a point is a subset that contains an open set - according to the topology. And moreover that open subset must contain the point. 🧐

Which open subsets are in the topology that contain 0? 🤔
 
Klaas van Aarsen said:
Nope. (Shake)

We can only have a ball with a radius if we can measure distances. But in a topology those are not defined. Instead a neighborhood of a point is a subset that contains an open set - according to the topology. And moreover that open subset must contain the point. 🧐

Which open subsets are in the topology that contain 0? 🤔

So do we consider an interval around $0$ ? :unsure:
 
  • #10
mathmari said:
So do we consider an interval around $0$ ?

No, we have to apply the definition of a cofinite topology.
It says that the open subsets are the empty set plus all subsets that have a complement that is finite. 🤔
 
  • #11
mathmari said:
So do we consider an interval around $0$ ?

Btw, after we've established what a neighborhood of $0$ is in the cofinite topology, we will look at an interval around $0$ that is inside the neighborhood. 🤔
 
  • #12
Klaas van Aarsen said:
at does an open neighborhood of $0$ look like? 🤔

Is it an open set that contains an open subset containing 0? :unsure:
 
  • #13
mathmari said:
Is it an open set that contains an open subset containing 0?
More precisely, it's a subset $V$ of $\mathbb R$ that includes an open set $U$ containing $0$.
Note that $V$ is not necessarily open.
$$0 \in U \subseteq V \subseteq \mathbb R$$
🧐

Now what was an open set in the cofinite topology of $\mathbb R$ again? 🤔
 
  • #14
Klaas van Aarsen said:
Now what was an open set in the cofinite topology of $\mathbb R$ again? 🤔

It is a set that has finite complement or is empty, right? :unsure:
 
  • #15
Let $X$ be an arbitrary set. The non-empty open nsets are the complements offinite sets. We have to define also the empty set.

A sequence $x_n \to x$ converges as for the cofinite topology iff for each open neighbourhood $U$ of $x$, $U = X \setminus \{s\}$ for a $s$ with $x \neq s$, it holds that almost all $x_n$ are in $U$. So if $x \neq s$, thenalmost all $x_n \neq s$. If infinitelymany $x_n=s$ then $x=s$.

So in this case:

Let $U$ be an open neighbourhood of $0$. Since $\Bbb R\setminus U$ must be finite, $U$ containsallbut a fininte numberof terms of the sequence. Therefore $x_n\rightarrow 0$.

Is that correct? :unsure:
 
  • #16
mathmari said:
Let $U$ be an open neighbourhood of $0$. Since $\Bbb R\setminus U$ must be finite, $U$ contains all but a finite number of terms of the sequence. Therefore $x_n\rightarrow 0$.

Is that correct?
Looks right to me. (Nod)
So we conclude that $0$ is a limit of the sequence.
Is it 'the' limit? Can we tell for instance whether the sequence converges to $1$? (Wondering)
 
  • #17
Klaas van Aarsen said:
Looks right to me. (Nod)
So we conclude that $0$ is a limit of the sequence.
Is it 'the' limit? Can we tell for instance whether the sequence converges to $1$? (Wondering)

The limit is all the numbers of the form $\frac{1}{n}$ with $n\in \mathbb{N}$, right? :unsure:
 
  • #18
mathmari said:
The limit is all the numbers of the form $\frac{1}{n}$ with $n\in \mathbb{N}$, right?

Suppose we pick a number that is not of that form. Let's say we pick $\pi$.
Then an open neighborhood $U$ of $\pi$ is all of $\mathbb R$ except for a finite number of points, and it must include $\pi$ itself.
The neighborhood $U$ will contain all but a finite number of the sequence won't it? 🤔
 

Similar threads

Replies
4
Views
2K
Replies
17
Views
969
Replies
61
Views
5K
Replies
21
Views
2K
Replies
1
Views
1K
Replies
9
Views
2K
Back
Top