# Analysis / upper lower bound

I am self studying Abbott's understanding analysis I have a question about two exercises.

Exercise: Give an example of each of the following, or state that the request is impossible:

1. A set $B$ with $\mathrm{inf}(B) \geq \mathrm{sup}(B)$.
2. A finite set that contains its infimum but not its supremum.

For 1. I said that this is impossible.
For 2. I understand that I must make an example of a set containing the greatest lower bound, but does not contain a least upper bound, but I don't know how to proceed.

Thanks.

• For the first one, think about a set with exactly one element, and remember, the infimum can equal the supremum. For the second one, think about why a finite set must contain it's supremum. – астон вілла олоф мэллбэрг Mar 29 '17 at 10:24

1) No that's not impossible. What you need is for a number $L$ to be both upper bound and lower bound of the set. That is for each $x\in B$ that $L\le x$ ($L$ is lower bound) and also $L\ge x$ ($L$ is upper bound). To state it differently it means that $L\le x \le L$ which means that $x=L$. Now there's only two possibility for $B$ left, ether $B=\{L\}$ or $B=\emptyset$. Now you calculate $\inf$ and $\sup$ of those and you'll see.
2) For a set $B$ to contain it's infimum $L$ we must have $L\in B$, but also that $L$ is a lower bound (that is there's no smaller $x$ in B). For $B$ to have a supremum $U$, means that there are members $x\in B$ such that $x\ge U-2^{-n}$, but also that there's no $x\in B$ larger than $U$. Now let $B=0$ and $U=1$ and you can construct such a set. It must certainly contain $0$ and if it contains $1-2^{-1}$ too it fulfills one of the requirements of $1$ being the $\sup$. The other is that none of the members may be $1$ or larger (which none of the mentioned is).