The discussion focuses on proving that the graph G of a continuous function f: X → Y is closed in the product space X x Y when Y is a Hausdorff space. Participants emphasize the need to show that the complement (X x Y) - G is open. A key approach involves selecting a point (x,y) in (X x Y) - G, where y ≠ f(x), and utilizing the Hausdorff property of Y to create disjoint neighborhoods around y and f(x). This leads to demonstrating the openness of the complement, thereby confirming the closed nature of G.
PREREQUISITESMathematicians, students of topology, and anyone interested in understanding the properties of continuous functions and their graphs in topological spaces.
Aureolux said:It can't be saying something like that, considering G is a subset of the product space (how can that be equal to a function?). All it means is any element of G [which, as a subset of the product space, has the form (x,y)] satisfies y=f(x), and any element in X x Y not in G does not satisfy this.
Although there are a few ways to show this, I will get you started on the way I think is most fundamental: assume there is an (x,y) in X x Y not in G. As I stated above this means y =/= f(x). Using the fact that Y is Hausdorff, you can form disjoint neighborhoods U and V around y and f(x), respectively. Try to use this to show that (X x Y) - G is open, and thus that G is closed. (Hint: we haven't used the continuity of f yet)
Damascus Road said:Let f: X \rightarrow Y be a function. The graph of f is a subset of X x Y given by G = {(x,f(x) | x \in X }. Show that if f is continuous and Y is Hausdorff, then G is closed in X x Y.
Any tips on how to start?
Is it saying that f: X \rightarrow Y = G ?