How would I figure out the glb and lub for the following.
$\{-2,2,-2.1,2.1,-2.11,2.11,...\}$
I would think the lub would be $2.12$
and the glb would be $-2$
And
$\{x:\ln(x)>0\}$
I am not sure how to find the glb and lub of this one.
$\endgroup$3 Answers
$\begingroup$What you'd actually have in the first case is the least upper bound as $2 +\dfrac 19 = \dfrac {19}{9}$, and greatest lower bound of $-2 - \dfrac 19 = -\dfrac {19}{9}$
In the second case, there is not any upper bound, let alone a least upper bound, as there is no upper bound for x for which $\ln x > 0$. Just keep making $x$ larger and larger. What can you say about the greatest lower bound? We know that "a" lower bound exists at $1$. Is this the greatest lower bound? (Recall that the lub and glb need not actually be in the given set.)
$\endgroup$ 13 $\begingroup$It's easy to see that any number from your first set is no larger than $2+\frac{1}{9}=2.\dot{1}$ and no smaller than $-2-\frac{1}{9}=-2.\dot{1}$.
For your second set, $\ln(x)>0$ iff $x>1$. So it is actually the set (interval) $(1,\infty)$.
So the infimum is $1$, and the supremum doesn't exist.
$\endgroup$ $\begingroup$$2.12$ doesn't work, because (for example) $2.115$ is also an upper bound. In fact, so are $2.1115,$ $2.11115,$ $2.111115,$ and so on. Consequently, if we have a number of the form
$$2.[\text{a bunch of }1\text{s in a row}][\text{digit}]$$
with $\text{digit}>1,$ then it will be an upper bound, but not the least upper bound. On the other hand, if we have a number of the form
$$2.[\text{a bunch of }1\text{s in a row}]0[\text{anything else}]\;,$$ then it isn't an upper bound at all. What is the only number that is viable as the least upper bound, then? Note the symmetry of the set about $0$ (that is, $x$ is in the set if and only if $-x$ is). From that, it follows that the glb will be the opposite (or negative) of the lub in this case.
For the second one, I recommend that you rewrite $\{x:\ln(x)>0\}$ as an interval. At that point it should be clear whether or not the set has an lub and/or a glb, and what they are.
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