Let $(x_n)$ and $(y_n)$ be bounded sequences such that $x_n ≤ y_n$ for all $n \in \mathbb{N}$. Show that $\limsup x_n ≤ \limsup y_n$ and $\liminf x_n ≤ \liminf y_n$.
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2Assume each statement is false and prove by contradiction. – lemon Sep 23 '14 at 19:25
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1Is this prove similar to proving Sup Xn≤ Sup Yn for the first part? – Gregory Eritsyan Sep 23 '14 at 19:28
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Use the proposition listed here and arrive at a contradiction after assuming the negation. – hrkrshnn Sep 27 '14 at 06:19
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Fix an index $k$. Then $$ m \ge k \implies x_m \le y_m \le \sup_{n \ge k} y_n.$$ Since this is true for all $m \ge k$ you have $$ \sup_{m \ge k} x_m \le \sup_{n \ge k} y_n.$$ Now take the limit on both sides as $k \to \infty$.
Umberto P.
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