2

Let $f:X \rightarrow Y$ continuous where the metric space $(X,d)$ is connected. On $Y$ we use the discrete metric. I want to show that $f$ must be constant on $X$. My approach:

We may assume that $|X| \geq 2$ (otherwise the claim is trivial). Let $x_0 \in X$. Assume $\exists y \in X: f(x) \neq f(y)$. Define $$ V:= \{ x \in X \mid f(x) = f(x_0) \} $$ and $$ W := \{x \in X \mid f(x) \neq f(x_0)\} $$ We have $V \cup W = X, V \cap W = \emptyset$ en $V,W \neq \emptyset$ by the assumption. I want to show that both $V,W$ are open to get a contradiction with the fact that $X$ is connected. By the continuity-property we know $$ \forall x \in X \exists \delta_x > 0 \forall y \in X : d(x,y) < \delta_x \rightarrow f(x) = f(y) $$ Let $x \in V$. Then $B(x,\delta_x) \subseteq V$ thus $V$ is open. Let $x \in W$. Then $f(x) \neq f(x_0)$. Further we have for all $y \in B(x,\delta_x)$ that $f(y) = f(x) $ and thus $f(y) \neq f(x_0)$ s.t. $B(x,\delta_x) \subseteq W$.

Is there an easier way to conclude this ?

Martin Argerami
  • 205,756

1 Answers1

3

You are making it too complicated, I think.

Suppose that $f$ is not constant. Then there exist $x_1,x_2\in X$ with $f(x_1)\ne f(x_2)$. Let $y_1 := f(x_1)$ and $y_2 := f(x_2)$. As $Y$ is discrete, the sets $\{y_1\}$ and $Y\setminus\{y_1\}$ are open (every subset of $Y$ is open!) and nonempty. Then $$ X=f^{-1}(\{y_1\})\cup f^{-1}(\{Y\setminus\{y_1\}) $$ is a union of disjoint open sets (open, because $f$ is continuous), a contradiction.

Martin Argerami
  • 205,756
  • 1
    I think you mean something like $\exists x_1,x_2 \in X: f(x_1) \neq f(x_2)$ where $y_1 = f(x_1)$ and $y_2 = f(x_2)$. The rest of the argument sounds very nice to me. Thanks for that. –  Jul 11 '13 at 17:25
  • You are right, of course. Thanks for the edit. – Martin Argerami Jul 11 '13 at 17:37