# The set of subsequential limits is closed

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.

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.

• Sir, I have confused with your notations. Can you explain with this notation.For $i=1,2,3,...$, considering the sequential a limit $y_{i}$, $\exists$ a subsequence ${x_{ni}}$ converging to $y_{i}$. (Since , each $y_{i}$ are the subsequential limit). I understand upto this. From this point how to choose wisely a subsequence of ${x_{n}}$ . How to choose the ${x_{ij}}$ which converges to y, from the given construction.
– user464147
Commented Jul 24, 2017 at 8:19
• @Mathdevotee : Did you understand the argument why for each $k$ there exist ${n_k}$ such that $|x_{n_k} - y_k| < 2^{-k}$? Commented Jul 25, 2017 at 12:59
• Here, I am confused with subscript. Can you explain ?. I understood like this first we choose $2^{-k}$[can I choose $\frac{1}{k}$ instead of that?] . Then we choose the $\epsilon$ definition of convergence of subsequence to $y_{k}$. Then move on toThe $\epsilon$ definition of convergence of subsequence to $y_{k+1}$[choosing $\epsilon$=$2^{k+1}$]. Procedure continues. Am I right?
– user464147
Commented Jul 25, 2017 at 17:44
• @Mathdevotee : Yes, you can think about it that way. Another way to think about this is to note that $y$ is a subsequential limit of $(x_n)_n$ iff for each $\varepsilon > 0$ and each $N$, there exists $n \geq N$ such that $|x_n - y| < \varepsilon$. For $y = y_k$ we take $\varepsilon = 2^{-k}$. And yes you can choose $\frac{1}{k}$ instead, the only important thing is that $|x_{n_k} - y_k|$ converges to zero for $k \to \infty$. Commented Jul 26, 2017 at 5:18
• I've added a proof in topological spaces, see my answer below. Commented Jul 8, 2020 at 12:09

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 $$k\geq p$$, $$x_{m_k}\in B_k\subset B_p\subset U.$$

• can you please elaborate on your point? Can you please write out a detailed proof of what you've stated? Commented Apr 19, 2016 at 13:29
• can someone please prove this claim? Commented Jul 7, 2020 at 19:37
• @SaaqibMahmood I've added a proof. Commented Jul 8, 2020 at 12:06
• @DavidWarrenKatz I've added a proof. Commented Jul 8, 2020 at 12:07
• so this only holds for first countable spaces? Commented Jul 9, 2020 at 14:58