# Convergence as for the cofinite topology on R

• MHB
Gold Member
MHB
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:

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MHB
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? 🤔

Gold Member
MHB
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:

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MHB
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)

Gold Member
MHB
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:

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MHB
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? 🤔

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MHB
What does an open neighborhood of $0$ look like? 🤔

Is it a ball with center the originand radius $\epsilon>0$ ? :unsure:

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MHB
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? 🤔

Gold Member
MHB
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:

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MHB
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. 🤔

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MHB
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. 🤔

Gold Member
MHB
at does an open neighborhood of $0$ look like? 🤔

Is it an open set that contains an open subset containing 0? :unsure:

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MHB
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? 🤔

Gold Member
MHB
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:

Gold Member
MHB
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:

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MHB
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)

Gold Member
MHB
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:

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MHB
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? 🤔