# Thread:

1. ## Question on monotone sequences

Let A be an infinite subset of the real numbers that is bounded above and let u=supA. Show that there exists an increasing sequence $(x_n)$ with $x_n \in A \forall n \in \mathbb{N}$ such that $u=lim(x_n)$.

The only way I can think of starting this question is to form some sort of sequence involving u, but can't think of how to do this. Help?

2. ## Re: Question on monotone sequences

Originally Posted by worc3247
Let A be an infinite subset of the real numbers that is bounded above and let u=supA. Show that there exists an increasing sequence $(x_n)$ with $x_n \in A \forall n \in \mathbb{N}$ such that $u=lim(x_n)$.

The only way I can think of starting this question is to form some sort of sequence involving u, but can't think of how to do this. Help?
Hint:

Use recursion.

Maybe I mistaken here, but here it goes:

Let us defined the following sequence:

$x_{n+1}=\sqrt{u+x_n}$, $x_1=\sqrt{u}\inA$( it because $A\neq\emptyset$).

Now, try to prove $x_n$ is monotonic increasing and bounded; and in the last stage find the limit.

3. ## Re: Question on monotone sequences

Originally Posted by Also sprach Zarathustra
$x_{n+1}=\sqrt{u+x_n}$, $x_1=\sqrt{u}\inA$( it because $A\neq\emptyset$).
How do you know that $x_n\in A, \forall n$?

@worc3247: use the definition of $\sup$: for all $\varepsilon>0$ we can find $x_{\varepsilon}\in A$ such that $u-\varepsilon<x_{\varepsilon}$. Now that $\varepsilon =\frac 1{n+1}$ for $n\in\mathbb{N}$.

4. ## Re: Question on monotone sequences

Originally Posted by worc3247
Let A be an infinite subset of the real numbers that is bounded above and let u=supA. Show that there exists an increasing sequence $(x_n)$ with $x_n \in A \forall n \in \mathbb{N}$ such that $u=lim(x_n)$.
That statement is false.
Let $A=[0,1]\cup\{2\}$.
There is no increasing (as opposed to non-decreasing) sequence in $A$ converging to $2$.

(I do not consider a constant sequence as increasing.)

5. ## Re: Question on monotone sequences

Well, it is. Otherwise, we have to say strictly increasing.

By the way, girdav's solution requires the axiom of countable choice. Maybe there is a way without it... I don't see it (except for a set with an ending interval).

6. ## Re: Question on monotone sequences

Originally Posted by pece
Well, it is. Otherwise, we have to say strictly increasing.
By the way, girdav's solution requires the axiom of countable choice. Maybe there is a way without it... I don't see it (except for a set with an ending interval).
I beg your pardon. There is a whole school of analysis for which that statement is false. There are just four types of monotone sequences: increasing, decreasing, non-increasing, or non-decreasing. When we see the word 'increasing' it means what it says, the sequence increases. If two consecutive terms are equal then there is no increase is there? So that sequence is possibly non-decreasing but it certainly not increasing.

7. ## Re: Question on monotone sequences

Sorry. For us french, increasing means non-decreasing for you. I translate it too literally.

In that case, it's clearly non-decreasing that would be used.

8. ## Re: Question on monotone sequences

The crucial point, as Plato said, is that you cannot prove the original statement- it is false.

If we are given that the sup of the set, u, is not in the set, then it is true. For any n> 0, there exist a member of the set in the interval from u- 1/n to u- otherwise u-1/n would be an upper bound on the set. Call that number " $x_n$". The (non-decreasing) sequence $\{x_n\}$ converges to u. A strictly increasing sequence can be derived from that sequence.
(We need u not in the set since otherwise, as in Plato's example, $x_n$ might be u itself.)

I have seen textbooks in English that use "non-decreasing; increasing" and others that use "increasing; strictly increasing". Normally which convention is being used is made clear.
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