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I consider the co-finite topology on $\mathbb{R}$ that is $$\{G\subset \mathbb{R}, card(\mathbb{R}\setminus G)<+\infty\}\cup \{\emptyset\}$$

What are the convergent sequences in this space .

thank you

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    A better question is "what are the limits of $(u_n)_n$?" or for what $x \in \mathbb{R}$ is the statement $u_n \rightarrow x$ true? – 2017-01-29

1 Answers 1

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Let $X$ be an infinite set and $(x_n)$ be a sequence in $X$ containing points that are all distinct (like the sequence $x_n = \frac{1}{n}$). Let $x \in X$ Then $x_n \rightarrow x$ in the cofinite topology on $X$.

Proof : let $O = X\setminus F$, where $F$ is finite, be an arbitrary non-empty open set that contains $x$. Then all $x_n$ are different by assumption, so let $N$ be the largest integer $n$ such that $x_n \in F$. (this number is then well-defined). Then for all $n > N$, $x_n \notin F$, so $x_n \in O$ for all such $n$. As $O$ was arbitrary, $x_n \rightarrow x$.

Also this is recommended reading for studying convergence in the co-finite topology.

So any injective sequence in the cofinite topology converges to every point of $X$.

In fact as the above link showed: the only convergent sequences are those that either

  1. take any value only finitely many times. So $\forall n \left|\{ m: a_m = a_n\}\right| < \infty$.

  2. have exactly one value that occurs infinitely many times: $\left|\{n: \left|\{m: a_n = a_m\}\right| = \infty \}\right| = 1$

Sequences from 1 include all injective sequences, the second class contains among others all eventually constant sequences.

Sequences of class 1 converge to any point of $X$, those of class 2 have the single infinitely-occurring value as their limit.

No other sequence is convergent.

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    Why you added " injective sequence" – 2017-01-29
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    @Vrouvrou to make the proof work (so there is a maximal $n$ with $x_n \in F$), and your example sequence obeys it, so it answers the question. Suppose that we had the sequence $1,2,1,2, \ldots$ then for $F = \{1\}$ we would have no such maximal $n$. There are some subtle cases with alternating sequences in the cofinite topology. The above one does not converge to $1$, as $1 \in O = X\setminus\{2\}$ shows, nor to $2$ as $2 \in O =X \setminus \{1\}$ shows, nor to any other point $x \neq 1,2$, as $O=X \setminus \{1,2\}$. Hence my restriction to "easy to prove" sequences. – 2017-01-29
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    @Vrouvrou it's a standard example to show that $x_n \rightarrow x$ and $x_n \rightarrow x'$ does not imply $x =x'$ as it does in nice spaces like Hausdorff spaces. Limits of sequences need not be unique, and can be very weird: in the cofinite topology on the reals we have $\frac{1}{n} \rightarrow \sqrt{2}$ being true as well as $\frac{1}{n} \rightarrow \pi$, e.g. – 2017-01-29
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    If i do the proof like this: suppose that anf sequence $(u_n)$ converge to $\ell$ an open containing $\ell$ is G such that $\mathbb{R}\setminus G$ is finit, then $(\mathbb{R}\setminus G)\cap \{u_n,n\in\mathbb{N}$ is finit and then there exists $n_0\in\mathbb{N}$ such that $\forall n\in \mathbb{N}, n\geq n_0; u_n\not\in \mathbb{R}\cap G $ this means that $\forall n\in \mathbb{N}, n\geq n_0; u_n\in G $ then every $\ell$ is a limit, where we must use "injectivity" of $(u_n)$ – 2017-01-29
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    @Vrouvrou $G$ is finite so $\{n: u_n \in G\}$ is finite: there are at most $|G|$ many such $n$ by $n \rightarrow u_n$ being injective. and a finite set has a maximum. – 2017-01-29
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    i don't understand , if $u_n$ is not injective , what hapen to $[\mathbb{R}\setminus G]\cap \{u_n,n\in \mathbb{N}\}$ – 2017-01-29
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    @Vrouvrou E.g. for the sequence $1,2,1,2\ldots$ and $G =\{1\}$, then $\mathbb{R}\setminus G$ contains infinitely many members of the sequence! Hence, the sequence does not converge to $2$! You must distinguish between the sequence which is a *function* $n \rightarrow u_n$, and its image, which is the set $\{u_n: n \in \mathbb{N}\}$. they are not the same. Convergence is about the function view, not the image set view! Beware... – 2017-01-29
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    eventualy constant sequence converge but they are not injective – 2017-01-31
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    @Vrouvrou this holds in all spaces. I did not write iff – 2017-01-31
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    so what is the iff condition – 2017-01-31
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    Please there two type of convergent sequence: the eventualy constant sequence which converge to a unique limite, and the injective sequences which converge to any limite . right ? – 2017-01-31
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    @Vrouvrou almost, I have a more accurate statement in my answer now. – 2017-01-31
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    you mean by |.| the card ? – 2017-01-31
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    @Vrouvrou yes, clear from the description, I would say – 2017-01-31