A set in the reals with a supremum and infimum Give an example or explain:
A set S $\subset \mathbb{R}$ with sup S $\in \mathbb{R}$ but inf S$\in${$-\infty , \infty$}
for some reason in my notes I explained a few months back:
S = $\{-n\}_{n=1}^{\infty}$
inf S = $-\infty$  sup S = -1 $\in \mathbb{R}$  
Is this correct? Also is there a more standard answer so that I may better conceptualize?
 A: Your example with $S= \{-n\}_{n=1}^\infty$ is correct. Having $\inf S=-\infty$ simply means $S$ is not bounded below, which is clearly the case since $S$ contains negative numbers of arbitrarily large magnitude. As for $\sup S=-1$, we need to check two conditions:


*

*$-1$ is an upper bound (i.e. $x \leq -1$ for all $x \in S$)

*$-1$ is a least upper bound (i.e. if $x \leq b$ for all $x \in S$, then $-1 \leq b$).


The first condition is clearly satisfied. For the second condition, suppose $b$ is another upper bound. Since $-1 \in S$, we must have $-1 \leq b$, as desired. Therefore $\sup S=-1$. 
In the above case, we could verify the second condition directly, but it is often easiest to approach it by contradiction. Also, there's no standard answer to "Find a set $S\subset \mathbb{R}$ with $\sup S\in \mathbb{R}$ but $\inf S \in \{-\infty,\infty\}$." We can extract some basic requirements though:  


*

*"$\inf S=\infty$" is usually taken to mean $S=\emptyset$. Observe that if $S$ is empty, then $\sup S=-\infty \not \in \mathbb{R}$. Therefore you can't have an example with $\sup S \in \mathbb{R}$ if $\inf S=\infty$. 

*The above constraint implies that we must have $\inf S = -\infty$, i.e. $S$ is not bounded below. It follows that $S$ must be infinite, but could be either countable or uncountable.

*$\sup S \in \mathbb{R}$ implies that $S$ is bounded above


Arguing in reverse, it is pretty easy to see that any nonempty set $S$ which is bounded above but not bounded below provides an example.
