Defining a function as transformation of open sets In a topological space $(X,\tau)$ is it true that a function $f$ is uniquely defined if we define $f$ as transformation of open sets in $X$?
My proof would be like follows (by contradiction)
Suppose $f,g$ are such that for each $\mathcal{O} \in \tau$ we have
$f(\mathcal{O}) = g(\mathcal{O})$, pick $x \in X$, and assume $f(x) \neq g(x)$. 
Let $\mathcal{O}_0 \subset \mathcal{O}$ an open set containing $x$ and pick $x_1 \neq x$ in $\mathcal{O}_0$. If we iterate we can construct a sequence of open sets $\mathcal{O}_k$ such that
$$
\begin{array}{l}
\mathcal{O}_k \subset\mathcal{O}_{k+1} \\
x \in \mathcal{O}_k \; \forall k \\
x_k \in O_{k-1} - \mathcal{O}_k
\end{array}
$$
such a sequence converges to $x$, if we assume that $f$ and $g$ are continuous then they will converge to the same point, since the for all $k$ we have $f(\mathcal{O}_k) = g(\mathcal{O}_k)$.
Is it correct?
 A: Suppose that $Y$ is a Hausdorff space.  Suppose that $f,g : X \to Y$ are continuous maps for which $f(\mathcal{O}) = g(\mathcal{O})$ for all open $\mathcal{O}$ in $X$.  We claim that $f=g$.
Suppose to the contrary that there is a point $a \in X$ with $f(a) \neq g(a)$.  As $Y$ is Hausdorff, we may separate these points:  take disjoint open sets $U$ and $V$ in $Y$ containing $f(a)$ and $g(a)$ respectively.  From $f(a) \in U$ we have $a \in f^{-1}(U)$, which is an open set.  Hence $g(a) \in g(f^{-1}(U))$.  But by hypothesis, this says $g(a) \in f(f^{-1}(U))$.  But $f(f^{-1}(U)) \subseteq U$, so we have $g(a) \in U$, which contradicts the disjointness of $U$ and $V$.  
Edit:  A similar argument works in the case that $Y$ is only $T_0$. The definition there is that $Y$ is $T_0$ if for all points $y_1 \neq y_2$ in $Y$ there is an open set $U$ containing one point but not the other (and you don't have control over which).  If we can enclose $f(a)$ in an open set $U$ not containing $g(a)$ then run the same argument above.  You obtain the same contradiction.  If not, then we can enclose $g(a)$ in an open set that doesn't contain $f(a)$ and you run the symmetric argument (flip the roles of $f$ and $g$) and still get the same contradiction.
