Let X and Y be topological spaces and assume that Y is Hausd

Let X and Y be topological spaces, and assume that Y is Hausdorff. Let f, g: X rightarrow Y be two continuous functions. Prove that {x X| f(x) = g(x)} is a closed subset of X.

Solution

2. X and Y are topological spaces and Y is Hausdorff.

Let f, g : X Y be two continuous functions.

To prove that {x X : f(x) = g(x)} is a closed subset of X.

Proof: Let A = {x X : f(x) g(x)}

Clearly A is a subset of X. We are intending to prove that A is an open subset of X.

Let x A, then f(x), g(x) Y such that f(x) g(x).

Since, Y is a Hausdorff space and f(x), g(x) Y such that f(x) g(x), there exists r > 0 such that

B_Y(f(x), r) B_Y(g(x), r) = {}

{} denotes null set, B_Y(f(x), r) denotes open ball in Y containing f(x).

B_Y(f(x), r) = {y Y : d_Y(f(x), y) < r}

Similarly B_Y(g(x), r) denotes open ball in Y containing g(x).

B_Y(g(x), r) = {y Y : d_Y(g(x), y) < r}

Since, f : X Y is a continuous function, x X and B_Y(f(x), r) be a neighbourhood of f(x), there is a neighbourhood U of x in A such that f(U) is contained in B_Y(f(x), r).

Similarly, g : X Y is a continuous function, x X and B_Y(g(x), r) be a neighbourhood of f(x), there is a neighbourhood V of x in A such that f(V) is contained in B_Y(g(x), r).   

Since, U and V are open sets in X containing x, their intersection is also open and x U V.

Therefore, there exists an open set U V in A containing x. This implies that A is open in X. Hence, the complement of A i.e. {x X : f(x) = g(x)} is closed in X. (Proved)