I want to prove following theorem
Let $S$ be the Sierpinski space and $Y$ a Hausdorff space. Show that every continuous map $f: S \rightarrow Y$ is a constant map.
First state my strategy. I am trying to show that continuous map is constant map. First what I know is the open set in Sierpinski space is follows : $T= \{ \{a\}, \{a,b\}, \phi\}$,
From the assumption, $Y$ is a Hausdorff, so for $x\neq y$, $\exists$ $U_x, V_y$ with $x\in U_x, y\in V_y, U_x\cap V_y = \phi$.
I tried to make up with $f^{-1}(U_x)$, $f^{-1}(V_y)$ is open and $f^{-1}(U_x \cap V_y) = f^{-1} (\phi)$ is open. But this is not enough for showing constancy of map $f$.
How I can show the constancy of map $f$?
Best Answer
Suppose that $f: S \to Y$ is not constant. This means that $f(a) \neq f(b)$, as $S$ only has the two points $\{a,b\}$. Now we need to apply Hausdorffness so there are open sets $U$ and $V$ in $Y$ such that
$$f(a) \in U, f(b) \in V, U \cap V = \emptyset\text{.}$$
This implies that $b \in f^{-1}[V]$ and this set is open, as $f$ is continuous, so $f^{-1}[V]=S$, as the only open set containing $b$ is $S$. But then $f(a) \in V$ too, which is a contradiction.
Generalisation: if $S$ is a space such that $$\forall U,V \text{ open }: U \neq \emptyset \land V \neq \emptyset \to U \cap V \neq \emptyset$$
(so all non-empty open sets intersect), then any continuous $f: S \to Y$ is constant when $Y$ is Hausdorff. The Sierpiński space obeys that property, among other spaces, like the cofinite topology on an infinite set, or the trivial topology.