I was trying to find an example of an oriented 2-surface S $\in {\Bbb R}^3$ such that at each point $p \in S$ , the principal curvatures of S at p are : $k_{1} (p) = 1, k_{2} (p) = -1$ .
My attempt:
We have a well-known result stating : On each compact oriented n-surface in ${\Bbb R}^{n+1}$,there exists a point p such that the second fundamental form at p is definite.
Now by the definition of the second fundamental form to be definite at some $p \in S$, it follows that either both the principal curvatures are positive (i.e. $k_{1} (p) , k_{2} (p) >0 $) or both the principal curvatures are negative (i.e. $k_{1} (p) , k_{2} (p) <0 $).
So given our conditions that $k_{1} (p) = 1, k_{2} (p) = -1$ ($\forall p \in S$), the contra-positive statement of the above mentioned result yields us that our required S cannot be compact . And since we are in $\Bbb R ^3$ i.e. in Euclidean space, S cannot be bounded. (Since S is always closed)
I could think up to this point but haven't been able to come up with a concrete example, thanks in advance for help.
This is an excellent question. You're right that such a surface, if it existed, would have to be noncompact. (And you can certainly find an example with those conditions at a single point $p$.)
There are two approaches here. One is to just work with the Codazzi-Mainardi equations and see whether such a surface can exist. More generally, you might try to classify non-umbilic surfaces first with one constant principal curvature, then with both principal curvatures constant. I refer you to exercises 16 and 17 on pp. 65-66 of my differential geometry text.