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Suppose $(a_1,\dots,a_n)$ is a sequence of real numbers such that $$a_1\leq a_2\leq \dots \leq a_n.$$

If $(b_1,\dots b_n)$ is a rearrangement of the sequence $(a_1,\dots,a_n)$ such that $$b_1\leq b_2\leq \dots \leq b_n,$$ then does it follow $a_1=b_1,\dots,a_n=b_n$?

I know that if the sequences were strictly monotonic, then the conclusion would have been obvious. How do I prove the question here, though... please help me with a proof maybe. Thank you in advance.

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  • $\begingroup$ Try with induction: Let X be the multiset of all numbers. $a_1$ and $b_1$ must both be the smallest element in X. Note that, in your case, $x\leq y$ and $y\leq x$ imply $x=y$. Thus, $a_1=b_1$. Now, remove one occurrence of $a_1=b_1$ from X, and continue by induction. $\endgroup$ – NeitherNor Jul 6 at 19:19
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Apart from the claim being obvious, Suppose $b_n\ne a_n$. If $b_n<a_n$, then $a_n=b_i$ for some $i<n$, but then $a_n=b_i\le \ldots \le b_n<a_n$, contradiction. Similarly, $b_n>a_n$ leads to a contradiction. Hence $a_n=b_n$. Now use induction on $n$.

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  • $\begingroup$ Thank you so much! $\endgroup$ – furfur Jul 6 at 19:15

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