Calculus Concepts and Modelling: An introduction to the concepts of calculus

Assume for the moment that $L<K\text{.}$ Take $b$ to be the average of $L$ and $K\text{.}$ Note that $b$ is strictly greater than $L$ and strictly less than $K\text{.}$ If $(s_n)\to L\text{,}$ then there is some tail ${(s_n)}_{n\ge N}$ completely contained in the open range $(L-1,b)$ around $L\text{.}$ But then no tail can be completely contained in the open range $(b,K+1)$ around $K\text{,}$ because this range is disjoint from the range around $L\text{.}$

In the case $L>K\text{,}$ make the same argument using range $(b,L+1)$ around $L$ and range $(K-1,b)$ around $K\text{.}$

Consider the contrapositive: If $(s_n)$ converged, then subsequences $(a_n)$ and $(b_n)$ would necessarily converge to the same limit value as $(s_n)$ (Pattern 22.3.13).

First, if there is any overlap between $(a_n)$ and $(b_n)\text{,}$ replace $(b_n)$ by a subsequence of itself with that overlap removed. If only a finite number of terms remain, that means that $(a_n)$ contains a full tail of $(s_n)$ and our conclusion follows from Fact 22.3.15. If an infinite number of terms remain in the new $(b_n)\text{,}$ then that sequence still converges to the same limit as before (Pattern 22.3.13) and we still have that $(a_n)$ and $(b_n)$ together account for all of the terms of $(s_n)\text{,}$ but now with no overlap.

Now let $L$ represent the common limit value of $(a_n)$ and $(b_n)\text{,}$ and let $(c,d)$ be some open range containing $L\text{.}$ Then there are tails ${(a_n)}_{n\ge N_1}$ and ${(b_n)}_{n\ge N_2}$ that are completely contained in $(c,d)\text{.}$ As $(a_n)$ and $(b_n)$ are subsequences, both $a_{N_1}$ and $b_{N_2}$ appear in $(s_n)\text{,}$ but they represent different terms in $(s_n)$ because we removed any overlap between $(a_n)$ and $(b_n)\text{.}$ Without loss of generality, assume that $a_{N_1}$ appears in $(s_n)$ before $b_{N_2}\text{,}$ and let $N$ be the index so that $b_{N_2}=s_N\text{.}$ (Note that this does not imply that $N_1<N_2\text{,}$ or that $N=N_2\text{.}$) Every term in the tail ${(s_n)}_{n\ge N}$ that appears in $(a_n)$ must do so with index greater than $N_1\text{,}$ since such a term will appear after $b_{N_2}\text{,}$ which appears after $a_{N_1}\text{.}$ And every term in the tail ${(s_n)}_{n\ge N}$ that appears in $(b_n)$ must obviously do so with index at least $N_2\text{.}$ In other words, every term in the tail ${(s_n)}_{n\ge N}$ is either in the tail ${(a_n)}_{n\ge N_1}$ or in the tail ${(b_n)}_{n\ge N_2}\text{,}$ each of which are completely contained in $(c,d)\text{.}$ Thus the tail ${(s_n)}_{n\ge N}$ is completely contained in $(c,d)\text{.}$

An unbounded sequence cannot be convergent, because if it were then it would have to be bounded by Fact 22.3.22.

We will consider only the case of a non-decreasing sequence; other cases are similar.

Suppose $(s_n)$ can be bounded within the closed range $R_0=[a,b]\text{.}$ We will create a convergent subsequence $(u_n)$ of $(s_n)\text{,}$ one term at a time. First, set $u_0=s_0\text{.}$

Split $R_0$ in half to create two subintervals. Since $R_0$ contains every of the infinite number of terms in $(s_n)\text{,}$ at least one of these two halves of $R_0$ must also contain an infinite number of terms. (But note that neither half necessarily has to contain a tail of $(s_n)\text{.}$) Let $R_1$ represent one of the halves of $R_0$ that contains an infinite number of terms, and set $u_1$ to be the term $s_{k_1}$ of $(s_n)$ of lowest index contained in $R_1\text{,}$ other than $s_0$ if that term happens to be $R_1\text{.}$

Now split $R_1$ in two halves, and let $R_2$ represent one of the halves that contains an infinite number of terms. Then set $u_2$ to be the term $s_{k_2}$ of $(s_n)$ of lowest index contained in $R_2\text{,}$ but so that the index $k_2$ is greater than the previous index $k_1\text{.}$

Finally, we may apply the Sequence Squeeze Theorem to conclude that the constructed subsequence $(u_n)$ converges to that common limit value of the two endpoint sequences.

Suppose it is true that every symmetric open range $(-c,c)$ containing zero contains some tail of the sequence. We must verify that it is then true that every open range $(a,b)$ containing zero, symmetric or not, contains some tail of the sequence. If this range contains zero, then necessarily $a<0$ and $b>0\text{.}$ Take $c$ to be the lesser of $\left|a\right|$ and $b\text{.}$ Then $c$ is a positive value and the range $(-c,c)$ is contained entirely within $(a,b)\text{,}$ so that if $(-c,c)$ contains a tail of the sequence, then $(a,b)$ contains that same tail.

Fact 22.3.31 implies that the statement is true in the case of $L=0\text{.}$ So assume $L\ne 0\text{,}$ and assume $(a,b)$ is an open range around $\left|L\right|\text{.}$ We may assume that $a\ge 0\text{,}$ since if we can determine a tail of $(\left|s_n\right|)$ that is contained in a smaller range with $a\ge 0\text{,}$ that same tail will also be contained in any larger range with $a<0\text{.}$

If $L>0$ then $(a,b)$ also contains $L\text{,}$ and so some tail ${(s_n)}_{n\ge N}$ is entirely contained in $(a,b)\text{.}$ But since $a\ge 0\text{,}$ $(a,b)$ is a range of positive numbers, and so ${(\left|s_n\right|)}_{n\ge N}$ and ${(s_n)}_{n\ge N}$ are identical sequences. Therefore, we may conclude that ${(\left|s_n\right|)}_{n\ge N}$ is entirely contained in $(a,b)\text{,}$ as required.