The set of subsequential limits is closed

Question

So I understand how to find subsequential limits of a sequence but why does the set of subsequential limits have to be closed? A line in my textbook briefly goes over this but doesn't really explain why. Any clarification? Thanks.

Answer 1

Let $(x_n)_{n \in \mathbb{N}}$ be a sequence and let $S$ be its set of subsequential limits. Let $y \in \overline{S}$ be an element in the closure of $S$, that is, there exists a sequence $(y_k)_{k \in \mathbb{N}}$ in $S$ converging to $y$. Since $y_k \in S$ is a subsequential limit there exists a number $n_k \in \mathbb{N}$ such that $|x_{n_k} - y_k| \leq 2^{-k}$ and for the same reason we can choose $(n_k)_{k \in \mathbb{N}}$ to be strictly increasing (if you like, by induction). Triangle inequality now gives $|x_{n_k} - y| \leq |x_{n_k}- y_k| + |y_{k} - y| $ for all $k \in \mathbb{N}$ and taking the limit shows that $(x_{n_k})_k$ is a subsequence of $(x_n)_n$ converging to $y$. Hence, $y \in S$ and so $S$ is closed.

Note: As you tagged your question "real-analysis", I used notation which gives the impression that this is happening over the real numbers, even though the same argument holds in any metric space. However, the argument does not work in arbitrary topological spaces but I believe that first-countability should be sufficient.

Answer 2

Claim: The set of subsequential limits can be written as $\displaystyle \bigcap_{n=0}^\infty \overline{\{x_{n+p}\mid p\in \mathbb N\}}$.

Proof:

• Let $\ell$ be a subsequential limit. There exists a strictly increasing sequence $(n_k)_k$ such that $\lim_k x_{n_k} =\ell$.
  Let $n\geq 0$. There is some $K\geq 0$ such that $n_K\geq n$. For $k\geq K$, $x_{n_k}\in \{x_{n+p}\mid p\in \mathbb N\}$ and $\lim_{k} x_{n_k} = \ell$. Thus $\ell \in \overline{\{x_{n+p}\mid p\in \mathbb N\}}$.

• Let $\ell \in \displaystyle \bigcap_{n=0}^\infty \overline{\{x_{n+p}\mid p\in \mathbb N\}}$. Consider two cases.
  First case: there are infinitely many $n$ such that $\ell \in \{x_{n+p}\mid p\in \mathbb N\}$. Then there is a subsequence $(x_{n_k})_k$ such that $\forall k, x_{n_k} = \ell$ and we are done.
  Second case: there are infinitely many $n$ such that $\ell \in \text{acc}(\{x_{n+p}\mid p\in \mathbb N\})$.
  Assume that there is a countable local basis at $\ell$ (which is the case when $(X,\mathcal T)$ is first-countable (e.g. when $X$ is a metric space)). Let $(B_n)_n$ denote a countable basis at $\ell$. WLOG $(B_n)$ is decreasing. Let $n_1$ be minimal such that $\ell \in \text{acc}(\{x_{n_1+p}\mid p\in \mathbb N\})$. There exists $m_1\geq n_1$ with $x_{m_1}\in B_1\setminus \{\ell\}$. There is some $n_2 > m_1$ with $\ell \in \text{acc}(\{x_{n_2+p}\mid p\in \mathbb N\})$. There exists $m_2\geq n_2$ with $x_{m_2}\in B_2\setminus \{\ell\}$, and so on. Then $x_{m_k}\to \ell$ as $k\to \infty$. Indeed, if $U$ is open with $\ell \in U$, there is some $p$ with $\ell \in B_p \subset U$. If $k\geq p$, $$x_{m_k}\in B_k\subset B_p\subset U.$$