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? – Henno Brandsma Jan 29 '17 at 12:02
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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
take any value only finitely many times. So $\forall n \left|\{ m: a_m = a_n\}\right| < \infty$.
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.
Henno Brandsma
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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. – Henno Brandsma Jan 29 '17 at 06:37
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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. – Henno Brandsma Jan 29 '17 at 06:40
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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 $(u_n)$ – Vrouvrou Jan 29 '17 at 11:27
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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. – Henno Brandsma Jan 29 '17 at 12:00
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i don't understand , if $u_n$ is not injective , what hapen to $[\mathbb{R}\setminus G]\cap {u_n,n\in \mathbb{N}}$ – Vrouvrou Jan 29 '17 at 12:04
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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... – Henno Brandsma Jan 29 '17 at 12:08
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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 ? – Vrouvrou Jan 31 '17 at 17:28
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@Vrouvrou almost, I have a more accurate statement in my answer now. – Henno Brandsma Jan 31 '17 at 18:15
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@Hennon please can we find the convergent sequences in the co-contable topology on R ? – Vrouvrou Jan 23 '19 at 20:13
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@Vrouvrou the only convergent sequences in the cocountable topology are the ones that are eventually constant. Lots of posts on that, some of them mine. – Henno Brandsma Jan 23 '19 at 20:15
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@Hennon if I let O an open set containing l, then $R\subset O$ must be all must countable, if it is finite then the injective sequences can converge to l, but if it is not finite we don't know ? – Vrouvrou Jan 23 '19 at 20:55
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@Vrouvrou If $x_n \to x$ define $F={x_n : x_n \neq x}$, which is countable so $x \in O= X\setminus F$ is open and so contains all $x_n$ from some $k$ onwards. But this means $x_n =x$ from $k$ onwards. – Henno Brandsma Jan 23 '19 at 21:52
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thank you very much please can you just tel me what means $\left|{n: \left|{m: a_n = a_m}\right| = \infty }\right| = 1$ in the co finite topology ? – Vrouvrou Jan 24 '19 at 16:20
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@Vrouvrou im my answer I already explained it: it means there is one value that appears infinitely many times as a term in the sequence. – Henno Brandsma Jan 24 '19 at 17:56
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What is the difference with the constant sequences $\exists n_0\in \mathbb{N}, \forall n\geq n_0: x_n=a$ – Vrouvrou Jan 24 '19 at 19:52
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@Vrouvrou eventually constant implies unique infinitely repeated value. – Henno Brandsma Jan 25 '19 at 18:00
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I think that in in "eventually constant" we can find the rang $n_0$ right ? – Vrouvrou Jan 25 '19 at 19:16
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