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I'm working on a homework assignment in which I have to find the upper and lower limits of a sequence. I've partitioned the sequence into two subsequences (one consisting of all even terms and another consisting of all odd terms), I've seen the following claim:

If we partition a sequence into a finite number of subsequences then the upper and lower limit of the sequence are equal to the maximum upper limit and minimum lower limit of the subsequences.

But we have never proved this in my class or textbook. How would I go about proving it for my assignment?

Thanks!

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  • $\begingroup$ It seems pretty trivial to me, since for a finite number of options $x_i$ we have $$\sup\{x_i\}=\max(\{x_i\})$$ $\endgroup$ – Uncountable Feb 21 '15 at 19:52
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The upper limit $s$ is the unique number satisfying the following two conditions:

(1) There is a subsequence converging to $s$.

(2) For all $s'>s$ there is no subsequence converging to $s'$.

Let's say you partitioned your sequence into $n$ subsequences and WLOG let the upper limit of the first subsequences $s_1$ be maximum. Then we show that $s_1$ satisfies the above two conditions.

(1) Just choose the subsequence to be the one in your subsequence $1$.

(2) For all $s'>s_1$, $s'$ is greater than the upper limits of all subsequences. Therefore there exist $N_1$, $N_2$, $\cdots$ such that $s'-\epsilon> a^{(1)}_n$ for all $n>N_1$, $s'-\epsilon> a^{(2)}_n$ for all $n>N_2$, etc. So $s'-\epsilon$ can be smaller than or equal to only a finite number of terms in the sequence, and so $s'$ cannot be the limit of any subsequece.

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