In the following (Terence 2006, p.38) I have to prove as exercice:
Let $f : X \to Y$ be a function from a connected metric space $(X, d)$ to a metric space $(Y, d_{disc} )$ with the discrete metric. Show that $f$ is continuous if and only if it is constant
I have proven this proposition without using the fact that $X$ is connected, so obviously I did something wrong. I would be very gratefull if you could point out my mistake.
Proof: $\impliedby$ Suppose $f$ is constant, i.e.,
$$\forall x,x'\in X: f(x) = f(x')$$
Choose an arbitrary $\varepsilon>0$. Then, no matter what our choice of $\delta_{\varepsilon}$ is, due to the definition of the discrete metric we always have
$$ d(x,x_0)<\delta_{\varepsilon} \implies d_{disc}(f(x),f(x_0))=0<\varepsilon$$
$\implies$ Now suppose that $f$ is continuous. Then, for any $x_0 \in x$ we have
$$\forall \varepsilon>0 \quad \exists \delta_{\varepsilon}>0 \quad \forall x \in X: \quad d(x,x_0)<\delta_{\varepsilon} \implies d_{disc}(f(x),f(x_0))<\varepsilon$$
Now suppose, for the sake of contradiction, that $f$ is not constant, i.e.,
$$\exists x_0 \in X \quad \forall x \in X/\{x_0\}: f(x) \neq f(x_0)$$
But then, for such $x_0$ we cannot find a $\delta_{0.5}$ so that $d(x,x_0)<\delta_{\varepsilon} \implies d_{disc}(f(x),f(x_0))<0.5$ – a contradiction.
Best Answer
Let $y_1,y_2 \in f(X)$. Write $Y=Y_1 \cup Y_2$ where $Y_1 \cap Y_2 = \emptyset$ and $y_1 \in Y_1,y_2 \in Y_2$. Since $Y_1$ and $Y_2$ are open sets as its a discrete metric space, $f^{-1}(Y_i)$ must be open and $X=f^{-1}(Y_1) \cup f^{-1}(Y_2)$ contradicting the connectedness of the metric space $X$.