all of the sets En is exactly the set of subsequential limits of the sequence f.

Question

Let $f : N \rightarrow R$ be a sequence and let $E_{n} = \{f(k) | k > n\}$. Prove of disprove that the intersection of the closures of all the sets $E_{n}$ is exactly the set of subsequential limits of the sequence $f$.

Answer 1

I assume that $R$ stands for "real numbers". Anyway everything that follows is true for any first-countable space.

Let $A$ be the set of all limits of all convergent subsequences of $(f_n)$ and let $B=\bigcap_m \overline{E_m}$. As usual you have to show two inclusions:

"$A \subseteq B$" Pick a convergent subsequence $(f_{n_k})$. By the definition of $E_m$ it follows that $(f_{n_k})$ eventually falls into $E_m$ regardless of $m$. Therefore $\lim_{k}f_{n_k}\in\overline{E_m}$ for each $m$. In particular $\lim_{k}f_{n_k}\in\bigcap_m\overline{E_m}$.

"$A \supseteq B$" Let $a\in\bigcap_m\overline{E_m}$. Since $a\in\overline{E_m}$ for each $m$ then it follows that $a=\lim a_n$ for some sequence $(a_n)\subseteq E_m$. The sequence $(a_n)$ depends on $m$ though, so lets denote it as $(a_n^m)$. Also note that it need not be a subsequence of $(f_n)$ even though it is composed of $f_n$'s. For all we know is that it can be for example constant.

So we get a sequence of sequences

$$\begin{matrix} (a_n^1)=& a_1^1 & a_2^1 & a_3^1 & a_4^1 & \cdots \\ (a_n^2)=& a_1^2 & a_2^2 & a_3^2 & a_4^2 & \cdots \\ (a_n^3)=& a_1^3 & a_2^3 & a_3^3 & a_4^3 & \cdots \\ (a_n^4)=& a_1^4 & a_2^4 & a_3^4 & a_4^4 & \cdots \\ \vdots& \vdots & \vdots & \vdots & \vdots & \ddots \end{matrix}$$

Now pick the diagonal: $b_n=a_n^n$. You can easily check that $b_n$ still converges to $a$. And by the definition of $E_m$ it follows that $b_m=f_k$ for some $k>m$. This property is enough to ensure that $b_m$ and $f_n$ have a common subsequence. This subsequence is what we were looking for and it completes the proof.