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Proposition 19(v) of Section 1.5 page 23 in the book Real Analysis by Royden and Fitzpatrick, 4th edition (see link), says:

If $a_n \le b_n$ for all $n$, then $\lim\sup a_n \le \lim \inf b_n$.

However one can find a counter-example for a sequence $a_n = b_n$ which has different limsup and liminf. Is it a mistake, or am I missing something obvious?

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  • $\begingroup$ Yes, this is an error. And your explanation of why it's wrong is exactly correct; I can't imagine why someone downvoted this. $\endgroup$ Sep 15, 2015 at 15:28
  • $\begingroup$ I guess you have to swap $\liminf$ with $\limsup$. $\endgroup$
    – egreg
    Sep 15, 2015 at 15:29
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    $\begingroup$ I didn't vote up or down, but perhaps the downvote is because the poster links to an unabridged pdf of an unauthorized reprint of a copyrighted book. MSE tends to frown on such shenanigans. $\endgroup$
    – Umberto P.
    Sep 15, 2015 at 15:37
  • $\begingroup$ @UmbertoP. That would be solved by an edit, not certainly a downvote. $\endgroup$
    – egreg
    Sep 15, 2015 at 15:39
  • $\begingroup$ Of course it would. $\endgroup$
    – Umberto P.
    Sep 15, 2015 at 15:39

3 Answers 3

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It seems you are right.

Take $a_n=0$ if $n$ is odd and $a_n=1$ if $n$ is even.

Take $b_n=a_n+0.5$

Then $\forall n$ $ a_n<b_n$ and $\limsup a_n=1$, $\liminf b_n=0.5$...

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You pinpointed the problem. Probably it's a typo and $$ \liminf\{a_n\}\le\limsup\{b_n\} $$ was meant.

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As others pointed out, it’s a typo, indeed.

What is true is that if $a_n\leq b_n$ termwise, then both $$\liminf_{n\to\infty}a_n\leq\liminf_{n\to\infty}b_n$$ and $$\limsup_{n\to\infty}a_n\leq\limsup_{n\to\infty}b_n$$ hold.

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