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Let $E$ be an uncountable set with $\tau:=$ co-countable topology.

WTP: A sequence in $E$ is convergent iff the sequence is eventually constant.

$"\Rightarrow"$ Assume $(x_n)_{n \in \Bbb N} \in E$ is a convergent sequence. So that is, $(x_n)_{n \in \Bbb N} \to l, l \in E$ Then, $\forall V \in \mathscr{N}_{\tau}(l), \exists N \in \Bbb N, n \ge N, x_n \in V$. Since $\tau$ is the co-countable topology $V^c$ is countable (or $V = \emptyset$). I can't figure out how to finish this one, more or less not sure how to put it down exactly. Seems the idea is that if the sequence is not uncountable at some point then it can't converge to something in the topological space.

$"\Leftarrow"$ Assume $(x_n)_{n \in \Bbb N}$ is eventually constant, then $\forall n \ge N$ we have $x_n \in \{x_N, x_{N+1}, \dots\} \wedge x_n = x_{N+i}, 0 \le i < \infty$ as $n \to \infty$. so $\forall V \in \mathscr{N}_{\tau}(x_N), \exists N \in \Bbb N,n \ge N, x_n \in V$ so it converges.

Martin Sleziak
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oliverjones
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2 Answers2

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Let $(x_n)_{n\in \mathbb{N}} \subseteq E$ be a convergent sequence and let $l\in E$ such that $x_n \rightarrow l$. Define $V:=E\setminus \{ x_n : x_n \neq l\}$. Clearly $V$ has countable complement and is therefore open in our topology. As $x_n \rightarrow l$ there exists $N\in \mathbb{N}$ such that for all $n\geq N$ holds $x_n \in V$. By definition of $V$ this means $x_n = l$ for all $n\geq N$, i.e. $(x_n)_{n\in \mathbb{N}}$ is eventually constant.

Let $(x_n)_{n\in \mathbb{N}} \subseteq E$ be eventually constant. Then there exists $l\in E$ and $N \in \mathbb{N}$ such that for all $n\geq N$ holds $x_n=l$. Let $V\subseteq E$ be an open neighborhood of $l$. By definition of neighborhood $l\in V$. Hence, for all $n\geq N$ holds $x_n\in V$. As $V$ was an arbitrary neighborhood of $l$ we conclude $x_n \rightarrow l$. Note that we didn't use any information about the topology, i.e. this holds in every topological space.

Severin Schraven
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Let E be a co-countable space and E be uncountable.

To show: A sequence in E is convergent iff the sequence is eventually constant.

If the sequence is eventually constant, then it converges to that constant.

Let $S = \{x_{n}\}_{n\geq 1} $ be a convergent sequence converges to $x$ in E.

Suppose S is not eventually constant, then for all $n\in \mathbb{N} $ there exists $m > n$ such that $x_{m} \neq x_{n}$.

Let n= 1, there exists $m_{1} > 1$ such that $x_{1} \neq x_{m_{1}}$ and for $m_{1}$ there exists $m_{2} > m_{1}$ such that $ x_{m_{1}} \neq x_{m_{2}} $ we can continue like this. So we get a subsequence $T =\{x_{m_{k}} \}_{k \geq 1} $ such that $ x_{m_{i}} \neq x_{m_{j}} $, if $i \neq j$.

Since T is a subsequence of S , hence converges to $x$.

Suppose x is not in T, then x is in complement of T which is a open set contradicts that T is a convergent sequence converging to x.

So, now let $x_{m_{l}} = x$ for some l. Now $x_{m_{l+k}}$ for all $l \geq 1$ is a subsequence of T which converges to x. Again by previous argument we can say that $ x_{m_{l+l_{0}}} = x$, for some $l_{0}$. This is a contradiction to the choice of T.

awCwa
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