# Show that all elements of one sequence are less than all elements of another sequence.

Let $$\{a_n\}_1^\infty$$ and $$\{b_n\}_1^\infty$$ be two sequences in $$\mathbb{R}$$ such that $$\forall n \in \mathbb{N}$$, it is true that $$a_n \leq b_n, a_n \leq a_{n+1}, \text{ and } b_{n+1} \leq b_n$$.

We want to show $$\forall m,n \in \mathbb{N}$$ it is true that $$a_m \leq a_n$$ and that there is a number $$r \in \mathbb{R}$$ such that $$a_m \leq r \leq b_n$$.

I've proceeding as follows:

We have $$a_{n} \leq b_{n} \implies a_{n+1} \leq b_{n+1}$$ and thus $$a_n \leq a_{n+1} \leq b_{n+1} \leq b_n$$.

Does this not imply that $$a_m \leq a_n$$? Even without stating the obvious fact that the sets are upper and lower bounds of each other? It seems then that $$r$$ would follow..

EDIT: Taking some of the ideas from below I have written a simple proof. Feedback is welcome and appreciated.

Since $$a$$ is monotonically increasing and $$b$$ is monotonically decreasing we have $$\forall m,n \in \mathbb{N}, a_n \leq a_{\max(m,n)} \leq b_{\max(m,n)} \leq b_n \implies a_m \leq b_n$$. Take $$r = a_{\max(m,n)} \text{ or } r = b_{\max(m,n)} \implies a_m \leq r \leq b_n$$.

It is clear that $a_n$ is a monotonically increasing sequence bounded above by $b_1$.Hence by Monotone Convergence Theorem $a_n\to r$ (say )

Since you have already proved that $a_n\leq b_n\forall n\in \mathbb N$ it follows that $r\leq b_n\forall n$

Hence $a_n\leq r\leq b_n$

• I've taken this idea and applied it in a slightly different way above. Oct 8 '15 at 14:31

First we note that $$b_1$$ is an upper bound for $$(a_n)$$. Since $$(a_n)$$ is monotone increasing, the Monotone Convergence Theorem states that $$(a_n)$$ converges to its supremum $$A$$. Similarly, $$a_1$$ is a lower bound for $$(b_n)$$. Since $$(b_n)$$ is monotone decreasing, the Monotone Convergence Theorem states that $$(b_n)$$ converges to its infimum $$B$$.

We claim that $$A\leq B$$. Suppose not. Then we have $$A>B$$. Let $$\varepsilon=\frac{A-B}{2}$$. Since $$A$$ is the limit of $$(a_n)$$, there exists $$M\in\mathbb{Z}$$ such that for all $$n\geq M$$, we have $$|a_n-A|<\varepsilon$$. Similarly, since $$B$$ is the limit of $$(b_n)$$, there exists $$N\in\mathbb{Z}$$ such that for all $$n\geq N$$, we have $$|b_n-N|<\varepsilon$$. Choose $$K=\max\{M,N\}$$. We notice that for all $$n\geq K$$, we have $$a_n>b_n$$, which contradicts $$a_n\leq b_n$$ for all $$n\geq 1$$.

Finally, for all $$n\geq 1$$, we have $$a_n < A \leq B < b_n$$.

You are asked to show $a_m \le r \le b_n$. Notice that the indices must be different, the result you give has the same index (i.e. $a_n \le r \le b_n$). Given $m,n \in \mathbb{N}$, we have without loss of generality $m\le n$. Then $a_m \le a_n \le b_n$ by your hypothesis, take $r=a_n$ and you are done.

• In my edit above I show something very similar written differently. Oct 8 '15 at 14:31