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Let $\left(x_{k_n^1}\right)_{n \geq 0}$ , $\left(x_{{k_n^2}}\right)_{n \geq 0}$ , $\dots$, $\left(x_{{k_n^t}}\right)_{n \geq 0}$ (where $t \in \mathbb{N^*}$) subsets of set $\left(x_n\right)_{n \geq 0}$ such that the sets $\left({k_n^1}\right)_{n \geq 0}, \left({k_n^2}\right)_{n \geq 0}, \dots, \left({k_n^t}\right)_{n \geq 0}$ form a partition of $\mathbb{N}$.

Show that $L(x_n) = L\left({k_n^1}\right) \cup L\left({k_n^2}\right)\cup \dots \cup L\left({k_n^t}\right)$, where $L\left({k_n^i}\right)$ is the set of limit points of the subsequence $\left(x_{k_n^i}\right)_{n\geq0}$.

What I've tried:

"$\supseteq$": Let $ x \in L\left({k_n^1}\right) \cup L\left({k_n^2}\right)\cup \dots \cup L({k_n^t}) \implies \exists i \in \{1,\dots,t\}$ such that $x \in L\left({k_n^i}\right) \implies \exists$ a subsequence $(y_n)_n$ of $\left(k_n^i\right)_n$ such that $y_n \stackrel{n}{\to} x$. But the subsequence $(y_n)_n$ is also a subsequence of $(x_n)_n$, so $x \in L(x_n)$.

"$\subseteq$": There i don't know how I can continue. Intuitively, think that if I take $x \in L(x_n)$, there exists a subsequence $(y_n)_n$ of $(x_n)_n$ s.t. $y_n \stackrel{n}{\to} x$. Because $(k_n^i), i=1,\dots,n$ form a partition we have countable many terms of $(x_n)$ in $\left(x_{k_n^1}\right)_n $ (chosen WLOG). So, we have a subsequence of $(y_n)_n$ in $\left(x_{k_n^1}\right)_n $ that has the limit in $L(k_n^1)$. Because $y_n \stackrel{n}{\to} x \implies x \in L(k_n^1) \implies x \in L({k_n^1}) \cup L({k_n^2})\cup \dots \cup L({k_n^t})$.

I don't know if the proof is good. Thanks!

Adam Rubinson
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MathLearner
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2 Answers2

3

It's fine, but personally, I would elaborate more on the argument and argue some parts slightly differently.

"$\subseteq$": Take $x \in L(x_n)$, then there exists a subsequence $(y_n)_n$ of $(x_n)_n$ s.t. $y_n \stackrel{n}{\to} x$.

If there does not exist $c\in\{1,\ldots, t\}$ such that $\left(x_{k_n^c}\right)_n $ contains infinitely many members of $(y_n),$ then this implies that for every $d\in\{1,\ldots, t\},\ \left(x_{k_n^d}\right)_n $ contains finitely many members of $(y_n)$. This would then imply that $(x_n) =\bigcup_{d\in\{1,\ldots,t\}} \left(x_{k_n^d}\right)_n\ $ contains only finitely many members of $(y_n),$ a contradiction. Therefore, our assumption that no such $c$ exists is false: there does exist $c\in\{1,\ldots, t\}$ such that $\left(x_{k_n^c}\right)_n $ contains infinitely many members of $(y_n).$ In other words, $\left(x_{k_n^c}\right)_n $ contains a subsequence of $(y_n).$ Call this subsequence (of $(y_n)_n$), $(y_{k_n})_n.$ Since $(y_{k_n})_n$ is a subsequence of $(y_n)$ and $y_n\to x,\ y_{k_n}\to x$ also.

We have shown that $\left(x_{k_n^c}\right)_n $ contains a subsequence, namely $(y_{k_n})_n,$ that converges to $x.$ Therefore, $x\in L\left(x_{k_n^c}\right)_n,\ $ and so $ x \in L({k_n^1}) \cup L({k_n^2})\cup \dots \cup L({k_n^t})$ also.

Adam Rubinson
  • 20,052
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The proof does hold. However, it is important to mention the fact that the result is dependent upon the fact that you have a finite number of $(k_n^i)_{n\geq 0}$. In order for you to select (WLOG) the sequence $(x_{k_n^1})_{n\geq 0}$ you must have that one of the sequences $(x_{k_n^i})_{n\geq 0}$ contains infinitely many terms of $(x_n)_{n\geq 0}$.

Nicholas White
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