Continuous Mappings into Hausdorff Space coinciding on Everywhere Dense Set coincide/Proof 1

Theorem

Let $\struct {X, \tau_X}$ be a topological space.

Let $\struct {Y, \tau_Y}$ be a Hausdorff space.

Let $f : X \to Y$ and $g : X \to Y$ be continuous mappings such that:

$\map f x = \map g x$ for all $x \in D$.

Then:

$\map f x = \map g x$ for all $x \in X$.

Aiming for a contradiction, suppose that there exists $x_0 \in X$ such that $\map f {x_0} \ne \map g {x_0}$.

Since $\struct {Y, \tau_Y}$ is Hausdorff, there exists an open neighborhood $U$ of $\map f {x_0}$ and an open neighborhood $V$ of $\map g {x_0}$ such that:

$U \cap V = \O$

We then have:

$x_0 \in f^{-1} \sqbrk U \cap g^{-1} \sqbrk V$

Let:

$W = f^{-1} \sqbrk U \cap g^{-1} \sqbrk V$

Since $f$ and $g$ are continuous, $W$ is the intersection of two open sets in $\struct {X, \tau_X}$.

So $W$ is open in $\struct {X, \tau_X}$ by the definition of a topology.

Since $D$ is an everywhere dense subset of $X$, there therefore exists $y \in D \cap W$.

Then we have $\map f y \in U$ and $\map g y \in V$.

Since $U \cap V = \O$, it follows that $\map f y \ne \map g y$.

But we also have $y \in D$, and by hypothesis we have that $\map f x = \map g x$ for all $x \in D$.

Hence we have a contradiction.

We conclude:

$\map f x = \map g x$ for all $x \in X$.

$\blacksquare$