# Continuity on hausdorff spaces

• May 18th 2009, 01:31 PM
Inti
Continuity on hausdorff spaces
Let $f$ and $g$ be two continous functions on a topological space $X$ with values on a hausdorff space $Y$. Prove that the set $\{x \in X : f(x)=g(x)\}$ is closed in $X$.

Any help is appreciated.
• May 18th 2009, 02:03 PM
Plato
Quote:

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

Suppose that $z\notin\{x \in X : f(x)=g(x)\}$.
Because Y is Hausdroff then $
\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 $f~\&~g$ are continuous then $z \in f^{ - 1} \left( {O_z } \right) \cap g^{ - 1} \left( {Q_z } \right)$ their intersection is open.
Clearly $t \in f^{ - 1} \left( {O_z } \right) \cap g^{ - 1} \left( {Q_z } \right)\; \Rightarrow \;f(t) \ne g(t)$.

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