Sequence Lemma and statistical convergence

Question

We know that from the sequence lemma : if a sequence $\{x_n\}$ in $A$ converges to $\ell$, then $\ell\in \bar A$. Conversely, if the space is first countable, then $\ell\in \bar A$ implies that $\exists$ a sequence $\{x_n\}$ in $A$ converges to $\ell$.

Also, every convergent sequence is statistically convergent sequence (and in this case limits & statistical limits are same). But the converse isn't true.

My question : If a sequence $\{x_n\}$ in $A$ converges statistically to $\ell$, does $\ell$ belong to $\bar A$?

Thanks in advance.

Answer 1

This is, in fact, quite trivial. Say there is a neighborhood $A=(\mathscr l-\epsilon, \mathscr l+\epsilon)$ of $\mathscr l$ such that $A$ contains no $x_n$. Then $A_\epsilon=\Bbb N$. Obviously, we have $\delta(\Bbb N)=1$. Thus, $\{x_n\}$ cannot converge statistically to $\mathscr l$.

Answer 2

So we have a "notion of smallness" $I$, an ideal on $\mathbb{N}$, i.e. a collection of subsets of $\mathbb{N}$ that contains all finite sets and is closed under shrinking ($A \in I, B \subseteq A \implies B \in I$) and closed under finite unions ($A, B \in I \implies A \cup B \in I$). We also demand that $\mathbb{N} \notin I$ to avoid trivialities.

It is a dual notion to a proper filter. In the paper this ideal is the sets of density $0$ (wrt the counting measure on $\mathbb{N}$), which clearly obeys all the ideal properties. In that case $x_n \to x$ wrt $I$ is called "statistical convergence". And $x_n \to x$ wrt an ideal $I$ (in the usual topology) iff

$$\forall \varepsilon >0: E((x_n), \varepsilon):=\{n \in \mathbb{N}: |x_n - x| \ge \varepsilon\} \in I$$

i.e. there are only "$I$-few" terms of the sequence outside the $\varepsilon$-ball around $x$ for any $\varepsilon$.

As finite sets are always in $I$, if $x_n \to x$ topologically, $x_n \to x$ for any ideal $I$.

Now suppose $x_n \to x$ for an ideal $I$ and all $x_n \in A$. If $B(x_n, \varepsilon)\cap A = \emptyset$ then $\mathbb{N}= E((x_n), \varepsilon)$, as $$n \in \mathbb{N}\rightarrow x_n \in A \rightarrow x_n \notin B(x, \varepsilon) \rightarrow |x_n -x| \ge \varepsilon \rightarrow n \in E((x_n), \varepsilon)$$

and this contradicts $x_n \to x$ under $I$ as $\mathbb{N} \notin I$.

So indeed $B(x,\varepsilon) \cap A \neq \emptyset$ and so $x \in \overline{A}$ as $\varepsilon >0$ was arbitrary. So under very mild conditions we indeed have the required property. All we need is convergence under an ideal (in fact it's much more common to consider the dual filter (the complements of the members of $I$) and talk about convergence along an ultrafilter. It comes down to the same thing.