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This is an exercise from an earlier calculus 1 reading at my university:

Let $X$ be a space containing infinitely many elements. In the cofinite topology, a set $\Omega$ is open iff $\Omega = \emptyset$ or $\Omega^c$ only contains finitely many elements. Which sequences converge and what is their limit?

As I had absolutely no idea how to approach this exercise, I looked up the solution and I found it quite frustrating that I didn't even understand the solution. The writer of the solution states that for any sequence $(a_n)$ in $X$, there are three possible cases:

a) There exists no value which the sequence takes infinitely many times.

b) There exists exactly one value which the sequence takes infinitely many times.

c) There exist two or more values which the sequence takes infinitely many times.

In the first case, the writer further states that the sequence converges to any value $a \in X$, in the second case, the sequence only converges to the value taken infinitely many times and in the last case, the sequence diverges. Obviously, the writer left out a proof or even a reasoning which might help a reader to understand why this is correct.

Can anyone help me out by explaining why this holds true? Thank you very much in advance.

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Do you know how the limit of a sequence is defined in topological spaces? – HbCwiRoJDp Nov 6 '11 at 19:40
I am not sure about this but I believe that a sequence converges to $a$, if for all neighbourhoods $U(a)$ there exists an integer $n_0 \in \mathbb{N}$ such that $a_n \in U(a)$ for all $n \geq n_0$. – Huy Nov 6 '11 at 19:46
If this question is really from a "calculus 1" course, I will have to remember that as we range over all the world's institutions "calculus 1" can mean almost anything. – Pete L. Clark Nov 6 '11 at 21:00
@PeteL.Clark: Yes, this question really is from a calculus 1 course. The professor holding this lecture usually includes a short "excursion" into topology (before talking about continuity) where he starts with easier concepts such as open, closed and compact sets and closes with relative topology and Hausdorff spaces. – Huy Nov 6 '11 at 21:55
up vote 17 down vote accepted

Case a: Let $(a_n)$ be a sequence in $X$ such that there exists no value which the sequence $(a_n)$ takes infinitely many times. Let $x \in X$ and let $U$ be an open set around $x$. Then $U^c$ is finite. Since $(a_n)$ does not take any value infinitely many times, after some index N, $(a_n) \in U$ for all $n \geq N$. Therefore $(a_n)$ converges to any $x\in X$. (This also proves that $X$ is not Hausdorff.)

Case b: Let $(b_n)$ be a sequence in $X$ such that there exists exactly one value $b \in X$ which the sequence $(b_n)$ takes infinitely many times. Let $U$ be an open set around $b$. Then $U^c$ is finite. Since $(b_n)$ takes at most finitely many values in $U^c$, $(b_n) \rightarrow b$. On the other hand, if $c \neq b$, $X \setminus \{b\}$ is an open set around $c$ that misses infinitely many $b_n$, so that the sequence cannot converge to $c$.

Case c: Let $(c_n)$ be a sequence in $X$ such that there exist two values $c_1$ and $c_2$ which the sequence $(c_n)$ takes infinitely many times. Let $U_1$ and $U_2$ be open sets around $c_1$ and $c_2$, respectively. Then $U_1^c$ and $U_2^c$ are finite by definition. Assume to the contrary, $c_n \rightarrow c$ for some $c \in X$. Let $V$ be an open neighborhood around $c$. Then there exists $N \in \mathbb{N}$ such that $(c_n) \in V$ for all $n \geq N.$ But this contradicts that $c_1$ and $c_2$ occurs in the sequence infinitely many times.

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For case a), we want to show that for any given $a$, and every open set containing $a$, that all the $a_n$ are in the open set as long as $n$ is greater than some $M$ which we get to choose. Since the open set in only missing a finite number of elements, we can put them in order $\{b_1, b_2, \ldots, b_p\}$. Then in the series$ a_n$, since $b_1$ does not appear infinitely many times, there is a last occurrence, call it $b_{1f}$. Similarly there is a last occurrence of $b_2$, call it $b_{2f}$. Then if $n \gt M=\max(b_{1f}, b_{2f},\ldots b_{pf})$, $a_n$ will be in our open set. So the sequence converges to $a$

The arguments for b) and c) are similar.

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Why is it that n $\gt$ M=$\max(b_{1f}$,$ b_{2f}$,$\ldots$ $b_{pf}$), shouldn't it be the sum of each $b_{pf}$ instead? – Arbër Ll Dec 15 '15 at 11:40
@ArbërLl: you could use the sum, or any number greater than the max, but the max is enough. We just need $n$ large enough that we are later in the sequence than any of the $b$'s. The maximum is the last appearance of one of them. – Ross Millikan Dec 15 '15 at 14:41
But for example if all $b_{pf}$ are equal to 1, then M=1, and for n=2, $a_2$ could be such a $b$ . Could you explain to me what am I doing wrong? – Arbër Ll Dec 19 '15 at 17:12
My $b_{if}$ are not the number of appearances of each element $b_i$, they are the last appearance of that element. If $b_{1f}=1000$, I am told that $b_1$ does not appear after position $1000$ in $a$. The $b_{if}$'s must all be distinct. If I am just given the count,I don't know how far out in $a$ I have to go to get past the last one. There might be just one occurrence of $b_1$, but it might be at position $1,000,000$. Then $b_{1f}=1,000,000$ – Ross Millikan Dec 19 '15 at 17:24

Say $X$ is a set endowed with the cofinite topology $\mathcal T$. Say $\{x_n\}_{n=1}^\infty$ is a sequence in $X$. We claim that $\{x_n\}$ converges in $(X,\mathcal T)$ if and only if there is at most one value in $\{x_n\}$ which occurs infinitely many times.

For the $(\Longrightarrow)$ direction, suppose that $\{x_n\}$ converges. Suppose $a$ and $b$ are distinct values in $\{x_n\}$ which occur infinitely many times. Observe that $X\setminus\{b\}$ is open and contains $a$. But there are infinitely many $x_n\notin X\setminus\{b\}$ and so $\{x_n\}$ cannot converge to $a$. Analogously, we see that $\{x_n\}$ cannot converge to $b$. But we assumed that $x_n\to c$ for some $c\in X$. Considering $X\setminus\{a,b\}$, we see by an analogous argument to that for $a$ and $b$, that $\{x_n\}$ cannot converge to $c$. Therefore, if $\{x_n\}$ converges, there is at most one infinitely repeating term of the sequence.

For the $(\Longleftarrow)$ direction, assume there is at most one infinitely repeating term of the sequence. There are two cases:

  1. There are no infinitely repeating terms. We claim that $\{x_n\}$ converges to every $x\in X$. Indeed, if $N(x)$ is a neighbhorhood of $x$, then $X\setminus N(x)$ is finite. As there are no infinitely occurring terms, there must be some $m\in\mathbb N$ such that $x_n\in N(x)$ whenever $n>m$.
  2. There is exactly one infinitely repeating term. Call this value $a$. We claim that $\{x_n\}$ converges to $a$ and only to $a$. If $N(a)$ is any neighborhood of $a$, then $X\setminus N(a)$ is finite. As no other term repeats infinitely many times, there is $m\in\mathbb N$ such that $m>n$ implies $x_n\in N(a)$. So $x_n\to a$. Now suppose $x_n\to a'$ for $a\neq a'$. Then $X\setminus\{a\}$ is open and contains $a'$, but there are infinitely many $x_n\notin X\setminus\{a\}$. So $\{x_n\}$ cannot converge to $a'$.
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