If $\limsup_{n\to\infty} \ x_{n} = a$ then why does it exist a subsquence $s_{n}$, which $\lim_{n\to\infty} \ s_{n} = a$? Maybe it seems trivial since $\limsup$ is known as the "greatest limit point of $x_n$", so there's a subsequence which converges to $a$. But I cannot use this definition. Is it possible to prove it without using this definition of $\limsup$? Thanks
 A: You want to prove

$\limsup\limits_{n\to\infty}x_n=a\;$ implies $\;a$ is a subsequential limit of $\{x_n\}$ 

See if you can get a contradiction from

$\limsup\limits_{n\to\infty}x_n=a\;$ and $\;a$ is not a subsequential limit of $\{x_n\}$

In particular, see if you can prove the following: If $a$ is not a subsequential limit of $\{x_n\},$ then there exists an open interval centered at $a$ that contains no terms of the sequence (or if you want, all you need to show is that there is an open interval containing $a$ that contains no terms of the sequence). From this result you should be able to obtain a contradiction to "$\limsup\limits_{n\to\infty}x_n=a$", regardless of what definition you use for $\limsup\limits_{n\to\infty}x_n.$ Regarding the "see if you can prove" part, note that if no such open interval exists, then you can pick terms from the sequence belonging to open intervals with center $a$ having arbitrarily small lengths, and these chosen terms will form a subsequence converging to $a.$
