A question that came up in the lecture of Friday June 16 was: why do we say functions are continuous if the inverse image of every open set is open? In particular why do we consider the inverse, rather than the forward image?
To gain intuition, consider functions f: R -> R with the usual topology on R. It turns out that even for a continuous function (in the normal calculus definition), the forward image of an open set need not be open. To see this, look at the function f(x)=x^2. Take the open interval (-1,1) in the domain of the function. Its image under f is the interval [0,1). This contains the point 0 and therefore is not an open set! The same can be true for the forward image of a discontinuous function such as f(x) = x+1, x < 0 and f(x)=x+2, x>=0, which is the example considered in class. The forward image of (-1,1), for example, is (0,1) U [2,3). Again this is not open (I may have said something wrong about this case on Friday). The message is that looking at the forward image of an open set tells us nothing useful about the continuity of a function R -> R.
The inverse image works much better. For any function R -> R that is continuous in the sense of calculus (both limits are equal at every point and are equal to the value of the function), the inverse image of an open set in the usual topology on R is always an open set. This is a theorem, the proof can be found in Section 1.4 of the book by Singer and Thorpe. You are encouraged to work it out.
Given the above results on continuous functions from R -> R, we abstract the property that the inverse image of open sets is always open, and use this to define continuity on an arbitrary topological space.
