# Thread: need help for proving

1. ## need help for proving

Let A be a nonempty subset of R that is bounded above. Show that there is a sequence $\displaystyle (x_n)$ of elements of A that converges to sup A.

Can we say that since A is bounded above, so is ($\displaystyle x_n$), then ($\displaystyle x_n$) is nondecreasing since it's bounded above, and so $\displaystyle x=$ sup{$\displaystyle x_n; n \in N$} exist.

2. ## Re: need help for proving

Originally Posted by wopashui
Let A be a nonempty subset of R that is bounded above. Show that there is a sequence $\displaystyle (x_n)$ of elements of A that converges to sup A.
Can we say that since A is bounded above, so is ($\displaystyle x_n$), then ($\displaystyle x_n$) is nondecreasing since it's bounded above, and so $\displaystyle x=$ sup{$\displaystyle x_n; n \in N$} exist.
Suppose that $\displaystyle \alpha=\sup(A)$.
If $\displaystyle \alpha\in A$ use a constant sequence consisting of $\displaystyle \alpha$.

If $\displaystyle \alpha\notin A$ then $\displaystyle \alpha-1<\alpha$ so
$\displaystyle \left( {\exists a_1 \in A} \right)\left[ {\alpha - 1 < a_1 < \alpha } \right]$
$\displaystyle \left( {\exists a_2 \in A} \right)\left[ {\max\{a_1,\alpha-2^{-1}\} < a_2 < \alpha } \right]$
For $\displaystyle n>2$ we have $\displaystyle \left( {\exists a_n \in A} \right)\left[ {\max\{a_{n-1},\alpha-n^{-1}\} < a_n < \alpha } \right]$

3. ## Re: need help for proving

Originally Posted by wopashui
Let A be a nonempty subset of R that is bounded above. Show that there is a sequence $\displaystyle (x_n)$ of elements of A that converges to sup A.

Can we say that since A is bounded above, so is ($\displaystyle x_n$), then ($\displaystyle x_n$) is nondecreasing since it's bounded above, and so $\displaystyle x=$ sup{$\displaystyle x_n; n \in N$} exist.
No, you can't say anything at all about $\displaystyle \{x_n\}$ until after you prove that it exists at which point you are done.

What you can do is assert that, with a= sup A, for any positive integer n, there exist a point of the set A in the interval (a, a+1/n). (Why?) Define $\displaystyle x_n$ to be that number.

4. ## Re: need help for proving

Originally Posted by HallsofIvy
What you can do is assert that, with a= sup A, for any positive integer n, there exist a point of the set A in the interval (a, a+1/n). (Why?)
I hope you meant "there exist a point of the set A in the interval (a-1/n,a)"?
Because $\displaystyle a=\sup(A)$ we have $\displaystyle A\cap (a,a+1/n)=\emptyset.$

If $\displaystyle a\in A$ and is an isolated point there is some k such that
$\displaystyle A\cap (a-1/k,a)=\emptyset.$

5. ## Re: need help for proving

Oh, yes, we are talking about sup, not inf. Thanks, Plato.

6. ## Re: need help for proving

So what Plato did is to prove the sequence $\displaystyle a_n = x_n$ exist, right, now we just need to show that it converges to supA?

7. ## Re: need help for proving

Originally Posted by wopashui
So what Plato did is to prove the sequence $\displaystyle a_n = x_n$ exist, right, now we just need to show that it converges to supA?
That is simple to do.
$\displaystyle \alpha - n^{ - 1} < x_n < \alpha$. BUT $\displaystyle \left( {\alpha - n^{ - 1} } \right) \to \alpha$

8. ## Re: need help for proving

Originally Posted by Plato
That is simple to do.
$\displaystyle \alpha - n^{ - 1} < x_n < \alpha$. BUT $\displaystyle \left( {\alpha - n^{ - 1} } \right) \to \alpha$
so by the squzze theorem, this is ture right?