(a) If the set $S$ of subsequential limits of $(x_n)$ is empty, then $(x_n)$ is unbounded.
(b) If $a$ is an eventual upper bound for $(x_n)$ and $b$ is an eventual upper bound for $(b_n)$, then $a-b$ is an eventual upper bound for $(a_n − b_n)$.
Using definitions of subsequential limits and eventual upper bounds, I need to show if these statements are true. Otherwise, I need to give a counterexample to disprove them.
I'm having a hard time with these. Here's what I'm thinking so far:
eventual upper bound is $limsup$ of $x_n$. By definition: $lim sup_{n→∞} x_n$ = inf{$β : ∃ N$ such that $x_n < β, ∀ n ≥ N$}.
a) If set $S$ of subsequential limits is empty, then the subsequences do not converge. By the Bolzano-Weierstrass theorem, every bounded sequence contains a convergent subsequence. By contrapositive, no convergent subsequences are contained in an unbounded sequence. So, $x_n$ unbounded. I can't come up with a counterexample for this, but I wasn't really sure.
b) $lim sup_{n→∞} a_n=a$ and $lim sup_{n→∞} b_n = b$. But does this always imply $lim sup_{n→∞} (a_n-b_n) = a-b$. I can't really come up with a counterexample for this, but I think it's true.
$lim sup_{n→∞} (a_n-b_n) = lim sup_{n→∞} a_n - lim sup_{n→∞} b_n = a-b$.
You have already answered first part. $(x_n)$ is necessarily unbounded, by Bolzano -Weierstrass Theorem.
Take $a_n=1$ and $b_n=-n$. Then $1$ is an essential upper bound for $(a_n)$ and $0$ is an essential uper bound for $(b_n)$ but $a_n-b_n$ is not bounded.