Question

Let A be a non-empty bounded subset of R with alpha=supA. Show that A contains a monotone increasing sequence with limit alpha

Answer 1

What course is this?

Answer 2

real analysis

Answer 3

@KatClaire Is there any sort of formula provided? And would you be able to expand on what alpha is in relation to your question?

Answer 4

like ok I know that if a sequence A is monotone increasing and alpha=supA you can show that alpha is the lim by
\[\epsilon > 0\] then there exists \[n _{o}\] such that \[\alpha - \epsilon < x_{n _{o}} \le \alpha \]
since \[x _{n} \ge x_{n_{o}} \forall n \ge n_{o}\] we have that \[\alpha - \epsilon < x_{n} \le \alpha \forall n \ge n_{o} \]
so \[\left| x_{n} - \alpha \right| < \epsilon \forall n \ge n_{o} \] HENCE..... \[\lim_{n \rightarrow \infty} x_{n} = \alpha\]

Answer 5

Here is the question

Answer 6

alpha would just be the upper bound of A @carson889

Answer 7

A = {x_n E R}: epsilon < x_n < alpha


We know that set A is bounded and thus it must have an upper bound and a least upper bound. We know the least upper bound (supremum) of A is alpha, which according to the least-upper-bound-property is both finite and exists.

Now, for every epsilon greater than zero, there exists an N such that x_N is greater than alpha - epsilon. This is necessary for alpha to be our limit or upper bound, otherwise alpha - epsilon would be our upper bound.

If A were to be increasing, then:

alpha - epsilon < x_N <= x_n < a < alpha + epsilon.

Using what we know:

- A is a subset of real numbers
- A is bounded
- alpha is sup(A) or the least upper limit of A, is the limit
- alpha is finite

We know there exists an x_N of subset A that is greater than alpha-epsilon, or else alpha would not be the upper bound, and lemma 1 of the monotone convergence theorem would not hold. We also know there exists an x_n greater than or equal to x_N and that x_n is less than the least upper bound

Since A is non-empty and we know that it is in part bounded above, then we know according to the least upper bound property of real numbers that alpha = sup(A) both exists and is non-infinite.

Example:

A = {x E R}: 3 <= x <= 20

lim = alpha = sup(A) = 20

epsilon = 3

alpha - epsilon = 17

Is there an x_N > alpha - epsilon ?

At x_16 > alpha - epsilon = 17

x_16 = 18. Thus 18 > 17

Now is there an x_n > x_N?

x_17 and x_18 = 19 and 20.

Therefore, there exists x_n > x_N.

From this we see that it must be increasing, and thus monotone with a limit of alpha = 20.

I hope this helps. I'll do my best to answer any questions, but frankly I have little experience in this subject.