Can a sequence have infinitely many subsequences that are disjoint?

Question

Can a sequence have infinitely many subsequences that are disjoint?

I wanted to prove the following,

Let $\{x_n\}$ be a sequence of a metric space with no converging subsequences. Then for each point outside the sequence $\{x_n\}$ has an open neighborhood that has no intersection with $\{x_n\}$.

Here is how I proceeded.

Let $\{ y_n \}$ be a subsequence and let $x$ be a point not in $x_n$. Since $y_n$ is not a convergent sequence it does not converge to $x$. Therefore, for all $N>0$, $\exists \epsilon>0$ such that $\{y_{n>N}\} \cap B(x,\epsilon) = \emptyset$. Since metric space is $T_0$ space, for any point $y_k \in \{y_{n}\} $ we can find $B(x,\epsilon_k)$ not containing $y_k$. Now the set $ U_y= \cap_{k\le N} B(x,\epsilon_k) \cap B(x,\epsilon) $ is an open set(intersection of finite number of open sets) with $U_y \cap \{y_n\} =\emptyset$.

Now consider the set $\{x_n\} - \{y_n\}$ and let $\{z_n\}$ be a subsequence of $\{x_n\} - \{y_n\}$. Then can find an open set $U_z$ such that $U_z \cap \{z_n\} =\emptyset$. Assume $\{x_n\}$ can only have a finite number of disjoint subsequences.

Then continue like that until we find finite number of points $S=\{x_n\} - \{y_n\}-\{z_n\} ...$. Again using the $T_0$ property, for each point $s \in S$, we can find an open ball around $x$, $B(x,\epsilon_s)$ that does not contain the point $s$. Therefore The set $U= \cap_{s\in S} B(x,\epsilon_s) \cap U_y \cap U_z \cap .. $ is a finite intersection of open sets that contain $x$ but does not contain $\{x_n\}$.

For the proof, I needed to use that there are only finitely many disjoint subsequences in $\{x_n\} $which I think should not be an assumption. Any comments or thoughts are really appreciated.

Answer 1

Consider the sequence

$$ 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, \ldots $$

Answer 2

Let $Y$ be the metric space, $X = \left\{x_n; n\in \mathbb N\right\}$. Assume the inverse is true: $$\exists y \in Y,\, y\not\in X,\, \forall r > 0,\, X \cap B\left(y, r\right)\neq \emptyset$$

Let's construct by induction a subsequence $\left(x_{n_k}\right)$ of $\left(x_n\right)$ such that $d\left(x_{n_k}, y\right) \le \frac1{k+1}$. Indeed, using $r=1$, we will have our $x_{n_0}$. Now assume that we construct $x_{n_0},\ldots,x_{n_k}$. Since $y\neq x_{\ell}$ for $\ell\in\{1,\ldots,n_k\}$, let $r = \min \left\{\frac1{k+2}, \frac12\min\left\{d\left(y, x_\ell\right): \ell=1,\ldots,n_k\right\}\right\}$ and choose $x_{n_{k+1}}$ subsequently. Since $d\left(x_{n_k}, y\right)\to 0$, this contradicts the assumption you have on $x_n$.

Answer 3

All sequences with infinitely many distinct elements have infinitely many subsequences that are distinct. Simply consider a subsequence that includes "half" of the members. (I put "half" in quotes because there are different ways of defining "half" of infinity. But it doesn't matter how you define it.) Now consider a subsequence that includes half of the remainder. Etc.

But you don't really need finitely many subsequences. Your problem is that you're just blindly taking a subsequence and showing that it doesn't converge to $x$, which is known as the "denying the antecedent" fallacy. If you're trying to prove $\neg(\exists X: P(X))$, it's not very productive to go through $X$ after $X$ proving that each one doesn't have property $P$, rather than providing a proof about all $X$ at once.

But consider what happens if this sequence is not finite. Then we have an infinite sequence $S=\{x_n\}−\{y_n\}−\{z_n\}...$ If we can force this subsequence to converge to $x$, then that contradicts the hypothesis that there are no convergent subsequences. And if we can show that a converging subsequence can be constructed any time $x$ fails to have an open neighborhood disjoint from $\{x_n\}$, then this means we have proven the required proposition (if $\neg x \rightarrow y$, then $\neg y \rightarrow x$).

So how do we force the subsequence to converge to $x$? Well, converging to $x$ means that for any $\epsilon$, we have something within $\epsilon$. So take $\{z_n\}$ to have half the $\epsilon$ of $\{y_n\}$, and take the next one have half the $\epsilon$ of that, and so on. If we can do this infinitely many times without getting an empty set, then taking an element from $\{y_n\}$, then taking an element from $\{z_n\}$, etc., is giving a subsequence converging to $x$.