Suprema and infima in real numbers I am having troubles when finding suprema and infima of sets. Could the supremum exist when the set is bounded from above? 
For example, $A=\{x\in \Bbb R \mid x^2<5\}$, the supremum is $\sqrt5$, the infimum is $-\sqrt5$ (is it right?). 
But for $B=\{x\in \Bbb Q \mid x^2<5\}$, what are the supremum and infimum? 
For $C=\{x\in \Bbb Z \mid x^2<5\}$, the supremum is $2$, the infimum is $-2$, and for $D=\{x\in \Bbb N \mid x^2<5\}$, the supremum is $2$, the infimum is 1. Are my answers correct for set $C$ and set $D$? (Note: $\Bbb N$ represents natural numbers and $0$ is not included.)
Moreover, what if we have a set $S=\bigcap_{n=1}^\infty ({-1\over n},{1\over n}]$ ? What are the supremum and infimum? The set is getting narrower as n goes to infinity, and I think $0$ could be neither supremum nor infimum.

For A, C, and D your answers are right (modulo Thomas Andrew's comment about $\Bbb{N}$.
Set $B$ has no $\sup x$ or $\inf x$.  That not-quite-trivial fact leads into the whole business of Dedekind cuts and putting the definition of the real line onto a rigorous footing.
Finally, since every $S_n =({-1\over n},{1\over n}]$ contains $0$, and for no other $x \in \Bbb{R}$ can we make that statement, $S = \bigcap_{n=1}^\infty S_n$ has zero as its supremum and its infinum.
