Ytukyg

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

$$
begin{array}{l}
mathcal{O}_k subsetmathcal{O}_{k+1} \
x in mathcal{O}_k ; forall k \
x_k in O_{k-1} - mathcal{O}_k
end{array}
$$

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?

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.