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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?

Ldog327
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1 Answers1

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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 $y_n = 1/n$.

kobe
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