$\{x_n\}$ be a bounded sequence such that for every bounded function $\{y_n\}$,
$$\limsup\{x_{n}+y_{n}\} =\limsup\{x_{n}\}+\limsup\{y_{n}\}$$ then prove that {$x_{n}$} is convergent. This is a question asked in in real analysis exam. I started answering like this : If $\limsup\{x_n\} = L$ then for every $\epsilon$ we can find natural number N such that $x_{n} <L+ \epsilon$ for n $\geq N$ A similar result holds for given bounded sequence $\{x_n\}$ but I cannot say that $L- \epsilon <x_{n}$, $n \geq N $.
Can we prove this result from the same definition of limit ? Or should we use some other results which connects limit superior and limit of a sequence? What is the importance of boundedness of the sequence (is it given to say that limsup is not $\infty$ ?)
Choosing $y_n=-2x_n$ we conclude from $$ \limsup_{n\to\infty} (x_n+y_n)=\limsup_{n\to\infty} x_n +\limsup_{n\to\infty} y_n $$ that $$ \limsup_{n\to\infty} (-x_n)=\limsup_{n\to\infty} x_n +\limsup_{n\to\infty}(-2x_n) $$ that is $$ -\liminf_{n\to\infty} x_n =\limsup_{n\to\infty} x_n -2\liminf_{n\to\infty} x_n $$ or equivalently $$ \liminf_{n\to\infty} x_n=\limsup_{n\to\infty} x_n. $$ and we are done. $\qquad\square$
Remark: We used strongly the fact that the sequence $\{x_n\}$ is bounded to insure that both $\liminf_{n\to\infty} x_n$ and $\limsup_{n\to\infty} x_n$ and we can do arithmetic operations on them. Consider what happens when $x_n=(-1)^n n$.