# Math Help - prove or disprove question

1. ## prove or disprove question

Let S be a non-empty set of real numbers and suppose that L is the least upper bound for S. Prove that there is a sequence of points $x_n$ in S such that $x_n$ --> L as n --> $\infty$.
Prove or disprove:
For non-empty bounded sets S and T: lub(SUT) = max{lub(S), lub(T)}.

2. I see that you have over fifty postings.
By now you should understand that this is not a homework service
So you need to either post some of your work on a problem or you need to explain what you do not understand about the question.

3. okay, here is what i got so far

for the first one, Let $x_0$ by any point of S. Since L is a least upper bound, for each n >0, there must be an element $(x_n)$ of S in the open interval ( $max(x_n-1,L-1/n),L$). Use these $x_n$.

For the second, i am thinking to prove by contradiction

4. For the first one there are two cases: i) $L\in S$ and ii) $L\notin S$.
In case i) take a constant sequence of L’s.
In case ii) there is a sequence of distinct terms such that $x_n \in \left( {L - \frac{1}{n},L} \right)$.

5. Originally Posted by Plato
For the first one there are two cases: i) $L\in S$ and ii) $L\notin S$.
In case i) take a constant sequence of L’s.
In case ii) there is a sequence of distinct terms such that $x_n \in \left( {L - \frac{1}{n},L} \right)$.
I think the question stated that we need to PROVE that there is a sequence of points $x_n$ such that $x_n\rightarrow L$ as $n\rightarrow\infty$.

I've done some work on this thanks to other sources, but I'm sure it either has some errors or is completely wrong.

For some $x_n\in S < L=lub(S), \exists x_{n+1}\in S = L$ such that $\frac{x_n+L}{2} where $x_n\rightarrow L$ as $n\rightarrow\infty$.

Can anyone tell me if I'm missing anything or if I did something wrong?

6. If $S=[0,1]\cup \{2\}$ then $L=2$.
The only sequence of points from $S$ converging to $2$ is come variation of $\left( {\forall n} \right)\left[ {a_n = 2} \right]$.

Does what you have work for this example?

7. Originally Posted by Plato
If $S=[0,1]\cup \{2\}$ then $L=2$.
The only sequence of points from $S$ converging to $2$ is come variation of $\left( {\forall n} \right)\left[ {a_n = 2} \right]$.

Does what you have work for this example?
I'm not exactly sure how that is meant to work into the first half of the question, or I'm mistaking it for being related to the first half when it's meant for the second.

I came up with my current answer from this link: http://www.mathisfunforum.com/viewtopic.php?id=1645, but I'm sure there's something in it that's missing.

8. The proof in the link is wrong. Use the same counterexample I gave.

The usual problem that goes with a similar proof is:
If $L=\text{LUB}(S)~\&~L\notin S$ then there is a sequence of distinct points from $S$ that converges to $L$.

But the example where $L\in S$ may not work except for almost constant sequence.