Greatest lower and leat upper bound help

• Dec 5th 2013, 01:01 AM
Tweety
Greatest lower and leat upper bound help
what is the least upper bound and greatest lower bound of the set

$\displaystyle [-1,3] \cap (\sqrt{2} , 5]$ ?

$\displaystyle { { x \in {R} : e^{x} \leq 2 } }$
• Dec 5th 2013, 03:42 AM
Melody2
Re: Greatest lower and leat upper bound help
Quote:

Originally Posted by Tweety
what is the least upper bound and greatest lower bound of the set

$\displaystyle [-1,3] \cap (\sqrt{2} , 5]$ ?

$\displaystyle { { x \in {R} : e^{x} \leq 2 } }$

I will start be admitting that I am new to this type of notation so if my answer sounds like nonsense then you are probably correct.

The intersection of those two intervals will be $\displaystyle (\sqrt2,3]$ I guess this is a limit on the x values

\displaystyle \begin{align*}e^x &\leq2\\lne^x & \leq ln2\\x&\leq ln2\\x&\leq 0.6931 approx\end{align*}

now if $\displaystyle x\leq ln2$ in the domain $\displaystyle (\sqrt{2} , 5]$ then there are no real solutions.
• Dec 5th 2013, 04:51 AM
romsek
Re: Greatest lower and leat upper bound help
Quote:

Originally Posted by Melody2
I will start be admitting that I am new to this type of notation so if my answer sounds like nonsense then you are probably correct.

The intersection of those two intervals will be $\displaystyle (\sqrt2,3]$ I guess this is a limit on the x values

\displaystyle \begin{align*}e^x &\leq2\\lne^x & \leq ln2\\x&\leq ln2\\x&\leq 0.6931 approx\end{align*}

now if $\displaystyle x\leq ln2$ in the domain $\displaystyle (\sqrt{2} , 5]$ then there are no real solutions.

I'm seeing 2 sets.

$\displaystyle \left.[-1,3]\cap \left(\sqrt{2},5\right.\right]$ and

{x : x is Real, ex <= 2}

OP doesn't specify any relation between these two sets so I think this is 2 problems.

$\displaystyle (\left.[-1,3]\cap \left(\sqrt{2},5\right.\right]$ = $\displaystyle \left.\left(\sqrt{2},3\right.\right]$ so

the LUB of the first set is clearly 3. The GLB is clearly $\displaystyle \sqrt{2}$

The LUB of the 2nd set = ln(2) as you noted. There is no lower bound on set 2 so there is no GLB.
• Dec 5th 2013, 07:28 AM
HallsofIvy
Re: Greatest lower and leat upper bound help
Yes, I agree that this is two separate problems.

MelodyII is right that the first set is equivalent to the interval $\displaystyle (\sqrt{2}, 3]$ so its least upper bound is 3 and greatest lower bound is $\displaystyle \sqrt{2}$.

romset is, of course, using the fact that the logarithm is an increasing function: if b $\displaystyle a\le b$ then $\displaystyle ln(a)\le ln(b)$

So if $\displaystyle e^x\le 2$ then $\displaystyle x\le ln(2)$. So there is no greatest upper bound on x and the least upper bound is ln(2).
• Dec 8th 2013, 04:08 AM
Melody2
Re: Greatest lower and leat upper bound help
why is it called the least upper bound and not just the upper bound?
and
why is it called the greatest lower bound and not just the lower bound?
• Dec 8th 2013, 05:29 AM
Plato
Re: Greatest lower and leat upper bound help
Quote:

Originally Posted by Melody2
why is it called the least upper bound and not just the upper bound?
and why is it called the greatest lower bound and not just the lower bound?

Definition: $\displaystyle L$ is the least upper bound of a set $\displaystyle A$ means that $\displaystyle L$ is an upper bound of $\displaystyle A$ such that if $\displaystyle K$ is also an upper bound of $\displaystyle A$ then $\displaystyle L\le K$.

Let $\displaystyle A=[0.8)$ then $\displaystyle 10$ is an upper bound of $\displaystyle A$ but $\displaystyle 8=\text{LUB}(A).$