# Math Help - Continuity, dense and Hausdorff

1. ## Continuity, dense and Hausdorff

Let f,g: X->Y be a continuous function. Assume that Y is Hausdorff and that there exists a dense subset D of X such that f(x)=g(x) for all $x \in D$ .

Prove that f(x)=g(x) for all $x \in X$

2. Originally Posted by Andreamet
Let f,g: X->Y be a continuous function. Assume that Y is Hausdorff and that there exists a dense subset D of X such that f(x)=g(x) for all $x \in D$ .

Prove that f(x)=g(x) for all $x \in X$
Since f(x)=g(x) for all $x \in D$ by the hypothesis of the problem, it remains to show that f(x)=g(x) for all $x \in X \setminus D$.

Suppose to the contrary that $f(x) \neq g(x)$ for some $x \in X \setminus D$. Since Y is Hausdorff, we have disjoint open sets U and V containing f(x) and g(x), respectively. Since f and g are continuous, x belongs to an open set $f^{-1}(U) \cap g^{-1}(V)$. We know that D is dense in X, which implies that every open set in X intersects a dense subset D. Thus, an open set $f^{-1}(U) \cap g^{-1}(V)$ intersects D. Let y belongs to that intersection. Since y belongs to $f^{-1}(U) \cap g^{-1}(V)$ , $f(y) \neq g(y)$. Since y also belongs to D, we have f(y)=g(y) by the hypothesis of the problem. Contradiction !

Thus, $f(x) = g(x)$ for all $x \in X \setminus D$.