Prove G is Closed: Continuous f, Hausdorff Y

[Damascus Road]

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 $$f: X \rightarrow Y = G$$ ?

 

[snipez90]

Um no, it's asking whether G is closed, so you start in the obvious manner: show (X x Y) - G is open. Pick a point (x,y) in (X x Y) - G and show that there exists a neighborhood of (x,y) that is contained in (X x Y) - G.

 

[Aureolux]

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]

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)

2 points x and f(x) and not just one (x,y) ?

 

[lavinia]

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 $$f: X \rightarrow Y = G$$ ?

For starter why not let X and Y be metric spaces? Then a Cauchy sequence inside the graph of f projects to Cauchy sequence in X and Y.