# Thread: Continuity on hausdorff spaces

1. ## Continuity on hausdorff spaces

Let $\displaystyle f$ and $\displaystyle g$ be two continous functions on a topological space $\displaystyle X$ with values on a hausdorff space $\displaystyle Y$. Prove that the set $\displaystyle \{x \in X : f(x)=g(x)\}$ is closed in $\displaystyle X$.

Any help is appreciated.

2. Originally Posted by Inti
Let $\displaystyle f$ and $\displaystyle g$ be two continous functions on a topological space $\displaystyle X$ with values on a Hausdorff space $\displaystyle Y$. Prove that the set $\displaystyle \{x \in X : f(x)=g(x)\}$ is closed in $\displaystyle X$.
Suppose that $\displaystyle z\notin\{x \in X : f(x)=g(x)\}$.
Because Y is Hausdroff then $\displaystyle \left( {\exists O_z } \right)\left[ {f(z) \in O_z } \right],\;\left( {\exists Q_z } \right)\left[ {g(z) \in Q_z } \right]\;\& \;O_z \cap Q_z = \emptyset$.

Because $\displaystyle f~\&~g$ are continuous then $\displaystyle z \in f^{ - 1} \left( {O_z } \right) \cap g^{ - 1} \left( {Q_z } \right)$ their intersection is open.
Clearly $\displaystyle t \in f^{ - 1} \left( {O_z } \right) \cap g^{ - 1} \left( {Q_z } \right)\; \Rightarrow \;f(t) \ne g(t)$.

Does this show that the complement of $\displaystyle \{x \in X : f(x)=g(x)\}$ is open?