TITLE: Supremums of sequences
QUESTION [0 upvotes]: Let $x_n$ and $y_n$ be two sequences of real numbers. Assume that $y_n$ is bounded above and that $x_n$$<$$y_n$ for all n$\in$$\mathbb{N}$. 
(a) Prove that $x_n$ is also bounded above. 
(b) Prove that sup($x_n$)$\leq$sup($y_n$). 
(c) Do the assumptions imply that sup($x_n$)$<$sup($y_n$)? If yes, prove it; if no, find a counterexample. 
I have proved (a). For part (b), I think I should try to prove by contradiction and suppose that sup($x_n$)$>$sup($y_n$)? Would this method work?

REPLY [1 votes]: Part (b) can be proven directly. For each index $k$, $x_k \le y_k \le \sup(y_n)$. So $\sup(y_n)$ is an upper bound for $\{x_k : k\in \Bbb N\}$. Therefore, $\sup(x_n) \le \sup(y_n)$. 
For part (c), consider $x_n = -1/n$ and $y_n = 1/n$.
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