Continuous mappings into a Hausdorff space

[radou]

Homework Statement

Here's another one I'd like to check.

Let f, g : X --> Y be continuous functions into a Hausdorff space Y. Show that S = {x in X : f(x) = g(x)} is closed in X.

The Attempt at a Solution

Let X\S be the complement of S in X. Let's show it's open.

Let x be an element of X\S = {x in X : f(x)$$\neq$$g(x)}. For f(x) and g(x), choose disjoint open neighborhoods U and V, respectively. Now, f^-1(U)$$\cap$$f^-1(V) is an open neighborhood of x which is contained in X\S. Since X\S can be written as a union of such open sets, it is itself open. Hence, S is closed.

 

[MathematicalPhysicist]

Let's take it step by step (like the series... (-:).

you take x in X\S, which means we have y1=f(x), y2=g(x) which are distinct, now you use Ys haussdorfian, and you take two sets U and V which are disjoint neighbourhoods of y1 and y2 respectively, x is contained in the intersection of f^-1(U) and g^-1(V) (and not both with f^-1), now because f and g are continuous f^-1(U) and g^-1(V) are open in X.
so you showed that X\S is contained in a union of the above open sets, now show the other inculsion, well it's kind of basic as well.

 

[radou]

Thanks for the reply.

I don't see why I should "use the other inclusion", since I managed to represent X\S as a union of its subsets "over all elements of X\S".

Yes, "f^-1" appeared twice - it was a mistype.