I think that $\limsup \big (-\infty, - n \sin \frac{1}{n} \big )$ (as the lim sup of sets) is $(-\infty, 1)$. Is this correct? If so, how do we prove it?
Background: I'm trying to do a question in Wheeden/Zygmund's Measure and Integral:
1.6) Compare $\displaystyle \limsup_{k\to \infty} a_k$ and $\limsup (-\infty, a_k)$.
Let $\displaystyle \limsup_{k\to \infty} a_k = a$. I am trying to come up with an example (if it is even possible) where $\limsup (-\infty, a_k) \neq (-\infty, a)$. Is this statement true? If so, are there other examples?
In general, for $a = \limsup a_n$, we can say the following: $$ (-\infty, a) \subseteq \limsup (-\infty, a_n) \subseteq (-\infty, a] $$ To see that, let $x < \limsup a_n$, then $a_n > x$ infinitely often, that is $x \in (-\infty, a_n)$ infinitely often, so $x\in \limsup (-\infty, a_n)$. On the other side, if $x > a$, there is an $N$ such that $x > a_n$ for $n \ge N$. That is $x \in (-\infty, a_n)$ at most for $n \ge N$, that is $x \not\in \limsup (-\infty, a_n)$.
We have $a \in \limsup (-\infty, a_n)$ iff $a_n > a$ infitely often, that is than depends on the particular sequence.