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.
I'm seeing 2 sets.
$\displaystyle \left.[-1,3]\cap \left(\sqrt{2},5\right.\right]$ and
{x : x is Real, e^{x} <= 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.
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).
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 $\displaystyle 8=\text{LUB}(A).$